Definition of the Derivative

Let’s delve into the “exciting” world of Calculus, starting with Differential Calculus! Think of it this way: Algebra helps us find the slopes of lines, and Differential Calculus helps us find the slopes of curves.

The derivative of a function is just the slope or rate of change of that function at that point. The reason we have to say “at that point” is because, unless a function is a line, a function will have many different slopes, depending on where you are on that function.

Why would we need to take a derivative in the real world? Let’s say an object was traveling along a curve, and we wanted to know how fast it was traveling (velocity) at certain points along that curve. If we had a function for the position of the object at certain times, we could take a derivative at certain points to know the velocity at that time. Velocity, then, is the rate of change or slope of position. By the same token, acceleration is the rate of change or slope of velocity. In fact, calculus grew from some problems that European mathematicians were working on during the seventeenth century: general slope, or tangent line problems, velocity and acceleration problems, minimum and maximum problems, and area problems.

The reason we need to know about Limits is because when we’re dealing with a curve, the actual slope of a part of the curve is constantly changing so theoretically we can’t actually take a derivative. We’ll zoom in on that part of the curve and use a limit to get the closest we can to the actual slope.

Tangent Line

To illustrate how we take slopes of curves, let’s draw a curve and illustrate the tangent line, which is a line that touches a curve at a certain (only one) point, and typically doesn’t go through that curve close to that point.

However, to get an actual slope of a line, we need two points instead of just one point. We must use what we call the secant line to define the slope (average rate of change), where this line goes through two other points on the curve. But we want this line to be tiny (so the slope is more accurate), so we want to use a limit where the change in $ \boldsymbol{x}$ gets closer and closer to $ 0$.

Here are some illustrations. Do you see how as we get smaller and smaller $ x$-values, there’s a much better chance the secant gets closer and closer to the actual tangent (slope) of the curve? Do you also see that as we get closer, the actual tangent line and secant lines become more and more parallel? This is what we want when we take the derivative in calculus: the tangent and secant lines basically become the same thing.

Definition of the Derivative

Here is the “official” definition of a derivative (slope of a curve at a certain point), where $ {f}’$ is a function of $ x$. This is also called Using the Limit Method to Take the Derivative.

Do you see how this is just basically the Slope of a Line formula (change of $ y$’s over change of $ x$’s)? Yes, $ \Delta x$ means “change in $ x$”, but for now,  think of it as another variable.

$ \displaystyle {f}’\left( x \right)=\underset{{\Lambda x\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+\Delta x} \right)-f\left( x \right)}}{{\Delta x}}$, provided the limit exists.

Note that just the quotient part of this formula (without the lim, or limit) is called the Differential Quotient. Don’t let this scare you away from Calculus! It’s really not that bad, and you actually won’t have to use this equation too often in Calculus. And note that not every function is differentiable, especially at certain points; for example, a function might be differentiable on an interval $ (a,b)$, but not at other points on its graph. For example, polynomials are typically differentiable, but rational functions are not at certain points (because of removable discontinuities and/or asymptotes).

Again, this derivative finds the slope of the tangent line to the graph of $ f$. It can also be used to find the instantaneous rate of change, or just rate of change, of one variable compared to another. And, as the $ x+\Delta x$ gets closer and closer to $ 0$, the average rate of change becomes the instantaneous rate of change.

To use this formula, we usually have to use the Limit Process that we learned about in the Limits section. The main thing we have to do is eliminate the $ \Delta x$ from the denominator since we can’t divide by $ 0$.

And just remember that for $ f\left( {x+\Delta x} \right)$, we just put  $ x+\Delta x$ everywhere where we have an $ x$ in the original function. (Note that I like to use the variable “$ h$” instead of $ ”\Delta x”$ since the algebra looks a little less messy). Here are some examples:

Definition of Derivative Problem Derivative:     $ \displaystyle {f}’\left( x \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)=3$

 

$ f\left( {x+h} \right)=3$    (since there’s no $ x$ on the right)

$ \require {cancel} \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\left( 3 \right)-\left( 3 \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( 0 \right)=0\end{align}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)=-3x$

 

$ f\left( {x+h} \right)=-3\left( {x+h} \right)=-3x-3h$

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{-3x-3h-\left( {-3x} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cancel{{-3x}}-3h-\cancel{{\left( {-3x} \right)}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{-3\cancel{h}}}{{\cancel{h}}}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {-3} \right)=-3\end{align}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)={{x}^{2}}-2x$

 

$ \displaystyle \begin{align}f\left( {x+h} \right)&={{\left( {x+h} \right)}^{2}}-2\left( {x+h} \right)\\&={{x}^{2}}+2xh+{{h}^{2}}-2x-2h\end{align}$

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{{{x}^{2}}+2xh+{{h}^{2}}-2x-2h-\left( {{{x}^{2}}-2x} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{{{{x}^{2}}}}+2xh+{{h}^{2}}\cancel{{-2x}}-2h\cancel{{-{{x}^{2}}}}\cancel{{+2x}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{2xh+{{h}^{2}}-2h}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{h}\left( {2x+h-2} \right)}}{{\cancel{h}}}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {2x+h-2} \right)=2x+0-2=2x-2\end{align}$

Here are a few more that are a little more complicated. Note that sometimes we have to find Common Denominators, and sometimes we have to use the trick where we Rationalize the Numerator by multiplying by a fraction with the Conjugate (changing the sign between terms) on the top and bottom. The last problem uses Trig Identities; note that there are other ways to do this using trig identities, but I found this is one of the simplest. Trust me; there are easier ways to take these derivatives!

Definition of Derivative Problem Derivative     $ \displaystyle {f}’\left( x \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}$
Use the Definition of the Derivative to find the derivative of:

$ \displaystyle f\left( x \right)=\frac{1}{{x+4}}$

 

 

$ \displaystyle f\left( {x+h} \right)=\frac{1}{{x+h+4}}$

$ \displaystyle \require {cancel} \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{1}{{x+h+4}}-\frac{1}{{x+4}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{{x+4}}{{\left( {x+h+4} \right)\left( {x+4} \right)}}-\frac{{x+h+4}}{{\left( {x+h+4} \right)\left( {x+4} \right)}}}}{h}\text{ }\,\,\,\,\text{ (common denominator)}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\frac{{\cancel{x}+\cancel{4}-\cancel{x}-h-\cancel{4}}}{{\left( {x+h+4} \right)\left( {x+4} \right)}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{-\cancel{h}}}{{\cancel{h}\left( {x+h+4} \right)\left( {x+4} \right)}}\,=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\left( {x+h+4} \right)\left( {x+4} \right)}}\,\\&=-\frac{1}{{{{{\left( {x+4} \right)}}^{2}}}}=-{{\left( {x+4} \right)}^{{-2}}}\end{align}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)=\sqrt{{x-2}}$

 

 

$ f\left( {x+h} \right)=\sqrt{{x+h-2}}$

 

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sqrt{{x+h-2}}-\sqrt{{x-2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sqrt{{x+h-2}}-\sqrt{{x-2}}}}{h}\cdot \frac{{\sqrt{{x+h-2}}+\sqrt{{x-2}}}}{{\sqrt{{x+h-2}}+\sqrt{{x-2}}}}\text{ }\,\,\text{ }(\text{use conjugate for diff. of squares})\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\left( {\cancel{x}+h-\cancel{2}} \right)-\left( {\cancel{x}-\cancel{2}} \right)}}{{h\left( {\sqrt{{x+h-2}}+\sqrt{{x-2}}} \right)}}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cancel{h}}}{{\cancel{h}\left( {\sqrt{{x+h-2}}+\sqrt{{x-2}}} \right)}}\\&=\frac{1}{{\left( {\sqrt{{x+0-2}}+\sqrt{{x-2}}} \right)}}=\frac{1}{{2\left( {\sqrt{{x-2}}} \right)}}=\frac{1}{2}{{\left( {x-2} \right)}^{{-\frac{1}{2}}}}\end{align}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)=2{{x}^{3}}-{{x}^{2}}+2$

 

 

$ f\left( {x+h} \right)=2{{\left( {x+h} \right)}^{2}}-{{\left( {x+h} \right)}^{2}}+2$

 

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{2{{{\left( {x+h} \right)}}^{3}}-{{{\left( {x+h} \right)}}^{2}}+2-\left( {2{{x}^{3}}-{{x}^{2}}+2} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{2\left( {x+h} \right)\left( {{{x}^{2}}+2xh+{{h}^{2}}} \right)-\left( {{{x}^{2}}+2xh+{{h}^{2}}} \right)+2-\left( {2{{x}^{3}}-{{x}^{2}}+2} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{2\left( {{{x}^{3}}+2{{x}^{2}}h+x{{h}^{2}}+{{x}^{2}}h+2x{{h}^{2}}+{{h}^{3}}} \right)-\left( {{{x}^{2}}+2xh+{{h}^{2}}} \right)+2-\left( {2{{x}^{3}}-{{x}^{2}}+2} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{{2{{x}^{3}}}}+4{{x}^{2}}h+2x{{h}^{2}}+2{{x}^{2}}h+4x{{h}^{2}}+2{{h}^{3}}\cancel{{-{{x}^{2}}}}-2xh-{{h}^{2}}\cancel{{+2}}\cancel{{-2{{x}^{3}}}}\cancel{{+{{x}^{2}}}}\cancel{{-2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{6{{x}^{2}}h+6x{{h}^{2}}+2{{h}^{3}}-2xh-{{h}^{2}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{h}\left( {6{{x}^{2}}+6xh+2{{h}^{2}}-2x-h} \right)}}{{\cancel{h}}}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\left( {6{{x}^{2}}+6xh+2{{h}^{2}}-2x-h} \right)=6{{x}^{2}}+6x\left( 0 \right)+2{{\left( 0 \right)}^{2}}-2x-0=6{{x}^{2}}-2x\end{align}$
Use the Definition of the Derivative to find the derivative of:

$ f\left( x \right)=\sin \left( {2x} \right)$

 

$ \begin{align}f\left( {x+h} \right)&=\sin \left[ {2\left( {x+h} \right)} \right]\\&=\sin \left( {2x+2h} \right)\end{align}$

 

Use these Special Trig Limits:

$ \displaystyle \begin{align}\underset{{x\to 0}}{\mathop{{\lim }}}\,&\frac{{\sin \left( x \right)}}{x}=1\\\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{1-\cos \left( x \right)}}{x}&=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \left( x \right)-1}}{x}=0\end{align}$

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2x+2h} \right)-\sin \left( {2x} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2x} \right)\cos \left( {2h} \right)+\cos \left( {2x} \right)\sin \left( {2h} \right)-\sin \left( {2x} \right)}}{h}\text{ }\,\end{align}$ (Identity: $ \sin \left( {x+y} \right)=\sin x\cos y+\cos x\sin y$)

 

$ \displaystyle \begin{align}&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2x} \right)\left[ {\cos \left( {2h} \right)-1} \right]+\cos \left( {2x} \right)\sin \left( {2h} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2x} \right)\left[ {\cos \left( {2h} \right)-1} \right]}}{h}+\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \left( {2x} \right)\sin \left( {2h} \right)}}{h}\,\,\,\left( {\text{Separate limits}} \right)\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2x} \right)\left[ {\cos \left( {2h} \right)-1} \right]}}{h}\cdot \frac{2}{2}+\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cos \left( {2x} \right)\sin \left( {2h} \right)}}{h}\cdot \frac{2}{2}\,\,\left( {\text{Multiply nums and dems by 2}} \right)\\&=2\sin \left( {2x} \right)\left[ {\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\left[ {\cos \left( {2h} \right)-1} \right]}}{{2h}}} \right]+2\cos \left( {2x} \right)\left[ {\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\sin \left( {2h} \right)}}{{2h}}} \right]\\&=2\sin \left( {2x} \right)\cdot 0+2\cos \left( {2x} \right)\cdot 1=2\cos \left( {2x} \right)\end{align}$

Equation of a Tangent Line

Note that there are more examples of finding the equation of a tangent line (including horizontal and vertical tangent lines) here in the Equation of a Tangent Line section.

Now that we know how to take the derivative (the more difficult way, at this point), we can also get the equation of the line that is tangent to a function at a certain point. This is because once we know the slope (derivative) of the curve at that point, we have a slope of a line, and a point on that line, so we can get the equation for the line.

When we get the derivative of a function, we’ll use the $ x$-value of the point given to get the actual slope at that point. Then we’ll use the $-y$ value of the point to get the complete line, using either the Point-Slope $ (y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right))$ or Slope-Intercept  $ (y=mx+b)$ method (for me, preferred). It’s really not too bad! (Weird fact: the equation of a tangent line for a linear function is just that function!)

Here are some examples. And I promise, taking the derivative will get easier when we learn all the tricks! Note that in the last problem, we are given a line parallel to the tangent line, so we need to work backwards to find the point of tangency, and then find the equation of the tangent line.

Function Equation of Tangent Line
Find the equation of the tangent line to the graph of $ f$ at the given point:

 

$ \begin{array}{c}f\left( x \right)=2{{x}^{2}}-2\\\,\text{Point:}\,\,\,\,\left( {-1,0} \right)\end{array}$

$ \displaystyle \begin{align}{f}’\left( x \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\left[ {2{{{\left( {x+h} \right)}}^{2}}-2} \right]-\left( {2{{x}^{2}}-2} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{{2{{x}^{2}}}}+4xh+2{{h}^{2}}\cancel{{-2}}-\cancel{{2{{x}^{2}}}}\cancel{{+2}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{4xh+2{{h}^{2}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{h}\left( {4x+2h} \right)}}{{\cancel{h}}}=4x+2\left( 0 \right)=4x\end{align}$

The slope of the function is $ 4x$, so at point $ \left( {-1,0} \right)$, the slope is $ m=4\left( {-1} \right)=-4$. Use either the slope-intercept or point-slope method to find the equation of the line (let’s use point-slope): $ y-0=-4\left( {x+1} \right);\,\,\,y=-4x-4$. Thus, the equation of the tangent line to $ f\left( x \right)=2{{x}^{2}}-2$ at point $ \left( {-1,0} \right)$ is $ y=-4x-4$.

Find the equation of the tangent line to the graph of $ f$ at the given $ x$ value:

 

$ \displaystyle f\left( x \right)=x+\frac{5}{x}$

$ \displaystyle x=0$

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\,\,\frac{{\left( {\cancel{x}+h} \right)+\frac{5}{{x+h}}-\left( {\cancel{x}+\frac{5}{x}} \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{h+\frac{5}{{x+h}}-\frac{5}{x}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{hx\left( {x+h} \right)+5x-5\left( {x+h} \right)}}{{h\left[ {x\left( {x+h} \right)} \right]}}=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{h{{x}^{2}}+{{h}^{2}}x+\cancel{{5x}}\cancel{{-5x}}-5h}}{{h\left[ {x\left( {x+h} \right)} \right]}}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{\cancel{h}\left( {{{x}^{2}}+{{h}^{2}}-5} \right)}}{{\cancel{h}\left[ {x\left( {x+h} \right)} \right]}}=\frac{{\left( {{{x}^{2}}+{{0}^{2}}-5} \right)}}{{x\left( {x+0} \right)}}=\frac{{{{x}^{2}}-\,5}}{{{{x}^{2}}}}\end{align}$

The slope of the function is $ \displaystyle \frac{{{{x}^{2}}-5}}{{{{x}^{2}}}}$, so when $ x=0$, the slope is undefined. Thus, no tangent line exists at $ \boldsymbol {x=0}$. (You can also see that the function doesn’t exist at this point, since there’s an asymptote at $ x=0$, so we really didn’t even need to take the derivative).

Find the equation of the tangent line to the graph of $ f$ and parallel to the given line:

 

$ \displaystyle f\left( x \right)=\frac{1}{{\sqrt{x}}}$

$ \displaystyle \text{Line:}\,\,\,\,2y+x=5$

$ \displaystyle \begin{align}{f}’\left( x \right)&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{f\left( {x+h} \right)-f\left( x \right)}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\frac{1}{{\sqrt{{x+h}}}}-\frac{1}{{\sqrt{x}}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\frac{{\sqrt{x}-\sqrt{{x+h}}}}{{\sqrt{{x\left( {x+h} \right)}}}}}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sqrt{x}-\sqrt{{x+h}}}}{{h\sqrt{{x\left( {x+h} \right)}}}}\cdot \frac{{\sqrt{x}+\sqrt{{x+h}}}}{{\sqrt{x}+\sqrt{{x+h}}}}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cancel{x}-\left( {\cancel{x}+h} \right)}}{{h\sqrt{{x\left( {x+h} \right)}}\left( {\sqrt{x}+\sqrt{{x+h}}} \right)}}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{-\cancel{h}}}{{\cancel{h}\sqrt{{x\left( {x+h} \right)}}\left( {\sqrt{x}+\sqrt{{x+h}}} \right)}}=\frac{{-1}}{{\sqrt{{x\left( {x+0} \right)}}\left( {\sqrt{x}+\sqrt{{x+0}}} \right)}}\\&=\frac{{-1}}{{\sqrt{{{{x}^{2}}}}\left( {2\sqrt{x}} \right)}}=-\frac{1}{{2x\sqrt{x}}}\,\,\,(=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}})\end{align}$

The slope of the tangent line is $ \displaystyle m=-\frac{1}{2}$ since the tangent line is parallel to $ \,2y+x=5$, or $ \displaystyle y=-\frac{1}{2}x+\frac{5}{2}$ (parallel lines have same slope). Set this slope to the derivative we found and solve for $ x$: $ \displaystyle -\frac{1}{2}=-\frac{1}{2}{{x}^{{-\frac{3}{2}}}};\,\,{{x}^{{-\frac{3}{2}}}}=1;\,\,{{\left( {{{x}^{{-\frac{3}{2}}}}} \right)}^{{-\frac{2}{3}}}}={{1}^{{-\frac{2}{3}}}};\,\,\,\,x=\frac{1}{{{{{\left( {\sqrt[3]{x}} \right)}}^{2}}}}=1$. Now plug $ x=1$ into the original function to get the $ y$-value for a point of tangency of $ (1,1)$. Use slope-intercept method to find the equation of the line: $ \displaystyle y=-\frac{1}{2}x+b;\,\,\,\,1=-\frac{1}{2}\left( 1 \right)+b;\,\,\,\,\,b=1+\frac{1}{2}=\frac{3}{2}$. Thus, the equation of the function’s tangent line that is parallel to the line $ 2y+x=5$ is $ \displaystyle y=-\frac{1}{2}x+\frac{3}{2}$. Tricky!

Definition of Derivative at a Point (Alternative Form of the Derivative)

If a derivative does exist at a certain point $ c$ (remember that it may not always), then we actually have an “easier” formula for this derivative (slope at this point). The cool thing is that again this looks just like a slope formula: change of $ y$’s over the change of $ x$’s:

$ \displaystyle {f}’\left( c \right)=\underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( c \right)}}{{x-c}}$

Here are some problems where we use this formula:

Function and $ \boldsymbol {x}$-value Alternate Form of the Derivative:   $ \displaystyle {f}’\left( c \right)=\underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( c \right)}}{{x-c}}$
Use the alternative form of the derivative to find the derivative at $ x=c$ (if it exists):

 

$ \begin{array}{c}f\left( x \right)=3{{x}^{3}}-1\\\,c=4\end{array}$

 

$ \displaystyle \begin{align}{f}’\left( 4 \right)&=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( 4 \right)}}{{x-4}}=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{\left( {3{{x}^{3}}-1} \right)-\left[ {3{{{\left( 4 \right)}}^{3}}-1} \right]}}{{x-4}}=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{3{{x}^{3}}-192}}{{x-4}}\\&=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{3\left( {{{x}^{3}}-64} \right)}}{{x-4}}=\underset{{x\to 4}}{\mathop{{\lim }}}\,\frac{{3\left( {\cancel{{x-4}}} \right)\left( {{{x}^{2}}+4x+16} \right)}}{{\cancel{{x-4}}}}\\&=3{{\left( 4 \right)}^{2}}+3\cdot 4\left( 4 \right)+3\cdot 16=144\end{align}$

(Note that we used a Difference of Cubes to factor the $ {{x}^{3}}-64$.) Thus, the derivative of $ f\left( x \right)=3{{x}^{3}}-1$ at $ x=4$ is 144. This is actually a little easier than using original method to find the derivative, which we could have done, plugging in $ x=4$ at the end.

Use the alternative form of the derivative to find the derivative at $ x=c$ (if it exists):

 

$ \begin{array}{c}f\left( x \right)=\sqrt{{x+4}}\\\,c=\,2\end{array}$

 

Use the conjugate to solve and rationalize the denominator at the end:

$ \displaystyle \require {cancel} \begin{align}{f}’\left( 2 \right)&=\underset{{x\to 2}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( 2 \right)}}{{x-2}}=\underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {\frac{{\sqrt{{x+4}}-\sqrt{{2+4}}}}{{x-2}}\cdot \frac{{\sqrt{{x+4}}+\sqrt{6}}}{{\sqrt{{x+4}}+\sqrt{6}}}} \right)\\&=\underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {\frac{{{{{\left( {\sqrt{{x+4}}} \right)}}^{2}}-{{{\left( {\sqrt{6}} \right)}}^{2}}}}{{\left( {x-2} \right)\left( {\sqrt{{x+4}}+\sqrt{6}} \right)}}} \right)=\underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {\frac{{\cancel{{x+4-6}}}}{{\cancel{{\left( {x-2} \right)}}\left( {\sqrt{{x+4}}+\sqrt{6}} \right)}}} \right)\\&=\underset{{x\to 2}}{\mathop{{\lim }}}\,\left( {\frac{1}{{\left( {\sqrt{{x+4}}+\sqrt{6}} \right)}}} \right)=\frac{1}{{\left( {\sqrt{{2+4}}+\sqrt{6}} \right)}}=\frac{1}{{\left( {2\sqrt{6}} \right)}}\cdot \frac{{\sqrt{6}}}{{\sqrt{6}}}=\frac{{\sqrt{6}}}{{12}}\end{align}$

Here’s one more that’s a little bit messy, since we’re dealing with another variable. (We used Difference of Cubes again.)

 

Use the alternative form of the derivative to find the derivative at $ x=a$ (if it exists):

$ f\left( x \right)=2{{x}^{3}}-{{x}^{2}}+2$

$ \displaystyle \begin{align}{f}’\left( a \right)&=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{2{{x}^{3}}-{{x}^{2}}+2-\left( {2{{a}^{3}}-{{a}^{2}}+2} \right)}}{{x-a}}\\&=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{2{{x}^{3}}-{{x}^{2}}+\cancel{2}-2{{a}^{3}}+{{a}^{2}}\cancel{{-2}}}}{{x-a}}=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{2{{x}^{3}}-2{{a}^{3}}-\left( {{{x}^{2}}-{{a}^{2}}} \right)}}{{x-a}}\\&=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{2\left( {{{x}^{3}}-{{a}^{3}}} \right)-\left( {{{x}^{2}}-{{a}^{2}}} \right)}}{{x-a}}=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{2\left( {x-a} \right)\left( {{{x}^{2}}+xa+{{a}^{2}}} \right)-\left[ {\left( {x-a} \right)\left( {x+a} \right)} \right]}}{{x-a}}\\&=\underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{\cancel{{\left( {x-a} \right)}}\left[ {2\left( {{{x}^{2}}+xa+{{a}^{2}}} \right)-\left( {x+a} \right)} \right]}}{{\cancel{{x-a}}}}=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {2{{x}^{2}}+2xa+2{{a}^{2}}-x-a} \right)\\&=2{{a}^{2}}+2a\left( a \right)+2{{a}^{2}}-a-a=6{{a}^{2}}-2a\end{align}$

Derivative Feature on a Graphing Calculator

You can use the nDeriv( (derivative) function on the TI graphing calculator to get the derivative (slope) of a function at a certain point;  hit math and then scroll down to nDeriv( or hit 8. To get the derivative at a certain point, put $ x$ in the denominator (after $ d$, for $ dx$) and put the value for $ c$ in at the end ($ x=c$). You can even graph the derivative of a function by using nDeriv (put $ x=x$ at the end) in the Y =  feature. Here are examples for the derivative of $ f\left( x \right)=3{{x}^{3}}-1$ at $ c=4$. 

You can also input a function and find the derivative at any point. After inputting and graphing the function (making sure the $ c/x$-value is in the window), use 2nd trace (calc) 6 ($ dy/dx$), hit ENTER, and type in $ c$ immediately (even though it doesn’t ask you for it; it will then say X = what you type, in our case, 4). We see that the derivative at that point is 144 again (you can ignore the Y-value).

You can also put a function in Y1 and put the derivative in Y2 by using nDeriv( with $ x$ after the “$ d$”, alpha trace enter (Y1 ) for the function, and then $ x$ at the end. Then you can see a function and its derivative on the same graph. Note that the derivative of a cubic function appears to be a quadratic.

Determining Differentiability

We learned above that not every function is differentiable at certain points (for examples, polynomials are differentiable at all points, while rational functions are not). In fact, the function may be continuous at a certain point, but not differentiable. (Note that the converse is true: if a function is differentiable at a point, it is also continuous at that point).

Here are some of the reasons that a function may not be differential at a point $ x=c$:

NOT Differentiable Example Graph   NOT Differentiable Example Graph
If it is not continuous at a point (example: $ x=1$).

 

Notes:

  • A point must have the same limit from the right and left to be differentiable.
  • The function IS differentiable at the endpoint, or when $ x=3$.
If there is a vertical tangent at that point (example: $ x=-1$).

 

Note that a horizontal tangent is differentiable.

If there is a sharp turn (a cusp or corner) at that point (example: $ x=0$). Note that the vertex on a parabola is differentiable, since the slopes are approaching $ 0$ from both sides.

 

Most piecewise functions aren’t differentiable at their boundary points. The exception to this is with a continuous piecewise function where the derivative is the same on either side of the “bump”; we’ll see an example below.

If there is a vertical asymptote at that point (example: $ x=2$).

 

Note that a horizontal asymptote is differentiable.

Derivatives from the Left and the Right

We can see that sometimes the derivative is different from the left and the right; in these cases, the function is not differentiable at the point where these derivatives are different.

Thus, in order for a function to be differentiable at a point $ \boldsymbol {x=c}$, $ \boldsymbol {f\left( x \right)}$ must be continuous at $ \boldsymbol {x=c}$, and $ \displaystyle \boldsymbol {\underset{{x\to {{c}^{-}}}}{\mathop{{\lim }}}\,{f}’\left( x \right)=\underset{{x\to {{c}^{+}}}}{\mathop{{\lim }}}\,{f}’\left( x \right)}$.

Here is an example:

Left and Right Derivative Problem Solution
For the function $ f\left( x \right)=\left| {x+1} \right|$, find the derivatives from the left and right at $ x=-1$.

 

 

Notice that since the derivatives are different from the right and the left at $ x=-1$, the function is not differentiable at that point.

This is an absolute value graph that is shifted to the left from the parent function:

Take the two derivatives at a point $ (-1)$, one from the left and from the right:

$ \displaystyle {f}’\left( {-1} \right)=\underset{{x\to -{{1}^{-}}}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( {-1} \right)}}{{x-\left( {-1} \right)}}=\underset{{x\to -{{1}^{-}}}}{\mathop{{\lim }}}\,\frac{{\left| {x+1} \right|-\left| {-1+1} \right|}}{{x-\left( {-1} \right)}}=\underset{{x\to -{{1}^{-}}}}{\mathop{{\lim }}}\,\frac{{\left| {x+1} \right|}}{{x+1}}=-1$

(since from the left, we’ll have $ (1-x)$ in the numerator)

$ \displaystyle {f}’\left( {-1} \right)=\underset{{x\to -{{1}^{+}}}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( {-1} \right)}}{{x-\left( {-1} \right)}}=\underset{{x\to -{{1}^{+}}}}{\mathop{{\lim }}}\,\frac{{\left| {x+1} \right|-\left| {-1+1} \right|}}{{x-\left( {-1} \right)}}=\underset{{x\to -{{1}^{+}}}}{\mathop{{\lim }}}\,\frac{{\left| {x+1} \right|}}{{x+1}}=1$

(since from the right, we’ll have $ (x-1)$ in the numerator)


Here are a few types of Piecewise Function problems you might see to make sure you understand if functions are continuous and/or differentiable. Note that we used shortcuts finding the derivatives that you’ll soon use in the Basic Differentiation Rules section.

Left and Right Derivative Problem Solution
For the function

 

$ \displaystyle f\left( x \right)=\left\{ \begin{array}{l}2x-2\text{ }\,\,\text{ }\,\,\text{ if }x\ge 2\\.5{{x}^{2}}\text{ }\,\,\,\,\,\,\,\,\,\,\,\text{ if }x<2\end{array} \right.$ ,

 

find the derivatives from the left and right at $ x=2$

 

Is this function continuous and/or differentiable?

This piecewise graph has a little “bump” at $ x=2$, at the boundary point between the intervals:

It’s a strange case where at $ x=2$, both the function and its derivative are the same from both sides. At $ x=2$, both parts of the piecewise function are the same:  $ 2(2)-2=.5{{(2)}^{2}}$. We will learn later that the derivative of $ 2x-2$ is $ 2$, and the derivative of $ .5{{x}^{2}}$ is $ x$, so at $ x=2$, both derivatives are $ 2$. Thus, the derivatives from the left and the right at $ x = 2$ are the same. Therefore, this piecewise function is both continuous and differentiable at $ x = 2$!

The function $ f$ is as follows:

 

$ \displaystyle f\left( x \right)=\left\{ \begin{array}{l}{{x}^{2}}+mx-2\text{ }\,\,\text{ if }x\le 1\\3x-b\text{ }\,\,\text{ }\,\,\,\,\,\text{ if }x>1\end{array} \right.$

 

For which values of $ m$ and $ b$ will the function $ f$ be both continuous and differentiable on its entire domain?

For the piecewise function to be continuous, both values need to be the same at $ x=1$, where the function changes course:

$ \begin{align}{{x}^{2}}+mx-2&=3x-b\\{{\left( 1 \right)}^{2}}+m\left( 1 \right)-2&=3\left( 1 \right)-b\\m&=4-b\end{align}$

Since we have two unknowns, we’ll need a System of Equations to solve. Since we also want the function to be differentiable on its entire domain, it needs to be differential at $ x=1$ (have same derivative from both sides). Take the derivative at both sides at this point, and make them equal:

$ \begin{align}2x+m&=3\\2\left( 1 \right)+m&=3\\m&=1\end{align}$

We now know $ m=1$ , and, from above, $ m=4-b$, so $ b=3$. Try it; it works!

Learn these rules, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Basic Differentiation Rules: Constant, Power, Product, Quotient and Trig Function Rules  – you are ready!