Definition of the Derivative

This section covers:

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 at points along 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) is called the Differential Quotient.

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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.

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 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.

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 rationals are not at every point (because of removable discontinuities and asymptotes).

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: $$f\left( x \right)=3$$   $$\begin{array}{c}f\left( x \right)=3\\f\left( {x+h} \right)=3\end{array}$$ (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: $$f\left( x \right)=-3x$$   \begin{align}f\left( x \right)&=-3x\\f\left( {x+h} \right)&=-3\left( {x+h} \right)=-3x-3h\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{{-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 \right)&={{x}^{2}}-2x\\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-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 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.

 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: $$\displaystyle f\left( x \right)=\frac{1}{{x+4}}$$     \begin{align}f\left( x \right)&=\frac{1}{{x+4}}\\f\left( {x+h} \right)&=\frac{1}{{x+h+4}}\end{align} \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{{\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{ (common denominator)}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\frac{{x+4-x-h-4}}{{\left( {x+h+4} \right)\left( {x+4} \right)}}}}{h}=\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: $$f\left( x \right)=\sqrt{{x-2}}$$     \begin{align}f\left( x \right)&=\sqrt{{x-2}}\\f\left( {x+h} \right)&=\sqrt{{x+h-2}}\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{{\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})\\&=\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: $$f\left( x \right)=\sin \left( {2x} \right)$$     \begin{align}f\left( x \right)&=\sin \left( {2x} \right)\\f\left( {x+h} \right)&=\sin \left( {2\left( {x+h} \right)} \right)\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}\text{ }\\&=\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}\end{align} Identify: $$\sin \left( {x+y} \right)=\sin x\cos y+\cos x\sin y$$   Trick: set $$\displaystyle \cos (2h)=1$$, as $$\displaystyle h\to 0$$, since $$\displaystyle \cos \left( 0 \right)=1$$.  (Can’t get rid of $$\displaystyle \sin (2h)$$ just yet, since the middle term would disappear): \displaystyle \begin{align}&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cancel{{\sin \left( {2x} \right)\cdot 1}}+\cos \left( {2x} \right)\sin \left( {2h} \right)-\cancel{{\sin \left( {2x} \right)}}}}{h}=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cos \left( {2x} \right)\sin \left( {2h} \right)}}{h}\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\cos \left( {2x} \right)\sin \left( {2h} \right)}}{h}\cdot \frac{2}{2}\,\,\,\,\,\,\left( {\text{Trick: multiply numerator and denominator by 2}} \right)\\&=\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{2\left( {\cos \left( {2x} \right)\sin \left( {2h} \right)} \right)}}{{2h}}=\cancel{{\underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( {2h} \right)}}{{2h}}}}\cdot \underset{{h\to 0}}{\mathop{{\lim }}}2\cos \left( {2x} \right)=2\cos \left( {2x} \right)\\\end{align} (Note that $$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\frac{{\sin \left( h \right)}}{h}=1$$.)

Equation of a Tangent Line

Note that there are more examples of finding the equation of a tangent line 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 or slope-intercept method. It’s really not too bad!

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:   $$\begin{array}{c}f\left( x \right)=2{{x}^{2}}-2\\\,\text{Point:}\,\,\,\,\left( {-1,0} \right)\end{array}$$ \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( {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} We know then the slope of the function is $$4x$$, so at point $$\left( {-1,0} \right)$$, the slope is $$m=4\left( {-1} \right)=-4$$. Now we can use either the slope-intercept or point-slope method to find the equation of the line (let’s use point-slope): $$y-0=-4\left( {x+1} \right);\,\,\,y=-4x-4$$. 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} We know then 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}}}\end{align} We know then the slope of the tangent line is $$\displaystyle m=-\frac{1}{2}$$ since the tangent line is parallel to $$\,2y+x=5$$ (parallel lines have same slope). We found the slope (derivative) of the function, so let’s solve for $$x$$: $$\displaystyle -\frac{1}{2}=-\frac{1}{{2x\sqrt{x}}};\,\,\,2x\sqrt{x}=2;\,\,\,x\sqrt{x}\,=1;\,\,\,\,{{\left( {x\sqrt{x}} \right)}^{2}}={{1}^{2}};\,\,\,\,{{x}^{3}}=1;\,\,\,\,x=1$$ When if we plug in $$x=1$$ in the original function, we see the point of tangency is $$(1,1)$$. Now let’s use the slope-intercept method to find the equation of the line: $$\displaystyle y=-\frac{1}{2}x+b;\,\,\,\,1=-\frac{1}{2}\left( 1 \right)+b;\,\,\,\,\,b=1+\frac{1}{2}=\frac{3}{2}$$. The equation of the function’s tangent line that is parallel to the line $$2y+x=5$$ is $$\displaystyle y=-\frac{1}{2}x+\frac{3}{2}$$.

Note that information about horizontal and vertical tangent lines can be found in the Equation of the Tangent Line section here

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:

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

Let’s do 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):   $$\begin{array}{c}f\left( x \right)=3{{x}^{3}}-1\\\,c=\,\,4\end{array}$$ \require {cancel} \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 had to use a difference of cubes to factor the $${{x}^{3}}-64$$.   Thus, the derivative of $$f\left( x \right)=3{{x}^{3}}-1$$ when $$x=4$$ is 144. Pretty cool! This is actually a little easier than using original method to find the derivative.   Note that if we had used the original definition of the derivative method, and put in 4 for $$x$$, we’d get this same slope! 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:   \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}

Derivative Feature on a Graphing Calculator

You can use the nDeriv( (derivative) function in your graphing calculator to get the derivative (slope) of a function at a certain point;  nDeriv can be found by hitting MATH and then scrolling down to nDeriv( or hitting 8.

Put $$x$$ in in the denominator (after $$d$$, for $$dx$$) and put your value for $$c$$ in at the end ($$x=c$$). Let’s get the derivative on a calculator for the first function above $$f\left( x \right)=3{{x}^{3}}-1$$  at $$c=4$$, as shown in the first display below.

You can even graph the derivative of a function by using nDeriv (put $$x=x$$ at the end) in the Y =  feature. (Note that the derivative of a cubic function appears to be a quadratic!):

Once you graph a function, you can also use 2nd trace (calc) 6 ($$dy/dx$$) to find the derivative of that function at a certain point $$c$$. Once you hit 6 and ENTER, you 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( and alpha trace enter for Y1 . Then you can see a function and its derivative on the same graph:

Determining Differentiability

We learned above that not every function is differentiable at certain points (for examples, polynomials are differentiable at all points, while rationals 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 If it is not continuous at a point (example: $$x=1$$):   Note that a point must have the same limit from the right and left to be differentiable.   Note that the function IS differentiable at the endpoint, or when $$x=3$$. If there is a sharp turn (a cusp or corner) at that point (example: $$x=0$$):   Note that a vertex, like on a parabola, is differentiable.   Most piecewise functions aren’t differentiable at their boundary points. Note: 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 tangent at that point (example: $$x=-1$$):   Note that a horizontal tangent is differentiable. 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.

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:   Let’s 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’s a piecewise function where the derivative is the same from the left and right; this function is differentiable:

 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 $$x = 2$$!

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