CALCULUS Quick Study Guide: DIFFERENTIATION  
DERIVATIVES:
Definition of Derivative: \(\displaystyle {f}’\left( x \right)=\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+h} \right)f\left( x \right)}}{h}\)
Derivative at a Point: \(\displaystyle {f}’\left( c \right)=\underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)f\left( c \right)}}{{xc}}\)
Constant Rule: \(\displaystyle \frac{d}{{dx}}\left( c \right)=0\)
Power Rule: \(\displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n1}}}\)
Product Rule: \(\displaystyle \frac{d}{{dx}}\left( {f\cdot g} \right)=f\cdot {g}’+g\cdot {f}’\)
Quotient Rule: \(\displaystyle \frac{d}{{dx}}\left( {\frac{f}{g}} \right)=\frac{{g\cdot {f}’f\cdot {g}’}}{{{{g}^{2}}}}\)
Chain Rule: \(\displaystyle \frac{d}{{dx}}\left( {f\left( u \right)} \right)={f}’\left( u \right)\cdot {u}’\)
Trig Derivatives: \(\displaystyle \begin{array}{l}\frac{d}{{dx}}\left( {\sin \left( x \right)} \right)=\cos \left( x \right)\\\frac{d}{{dx}}\left( {\cos \left( x \right)} \right)=\sin \left( x \right)\\\frac{d}{{dx}}\left( {\tan \left( x \right)} \right)={{\sec }^{2}}\left( x \right)\\\frac{d}{{dx}}\left( {\cot \left( x \right)} \right)={{\csc }^{2}}\left( x \right)\\\frac{d}{{dx}}\left( {\sec \left( x \right)} \right)=\sec \left( x \right)\tan \left( x \right)\\\frac{d}{{dx}}\left( {\csc \left( x \right)} \right)=\csc \left( x \right)\cot \left( x \right)\end{array}\)
(Trig functions starting with “c” are negative. I also remember that there are always 2 sec’s (csc’s) and 1 tan (cot’s) in the last 4 equations.)  Inverse Trig Derivatives: \(\displaystyle \begin{align}\frac{{d\left( {\arcsin u} \right)}}{{dx}}&=\frac{{{u}’}}{{\sqrt{{1{{u}^{2}}}}}}\\\frac{{d\left( {\arctan u} \right)}}{{dx}}&=\frac{{{u}’}}{{1+{{u}^{2}}}}\\\frac{{d\left( {\text{arcsec}\,u} \right)}}{{dx}}&=\frac{{{u}’}}{{\left u \right\sqrt{{{{u}^{2}}1}}}}\\\frac{{d\left( {\arccos u} \right)}}{{dx}}&=\frac{{{u}’}}{{\sqrt{{1{{u}^{2}}}}}}\\\,\frac{{d\left( {\text{arccsc}\,u} \right)}}{{dx}}&=\frac{{{u}’}}{{\left u \right\sqrt{{{{u}^{2}}1}}}}\\\frac{{d\left( {\text{arccot}\,u} \right)}}{{dx}}&=\frac{{{u}’}}{{1+{{u}^{2}}}}\end{align}\)
Exponential and Log Derivatives: (u is function of x, a is constant) \(\displaystyle \begin{align}\frac{d}{{dx}}\left( {\ln u} \right)&=\frac{{{u}’}}{u}\\\frac{d}{{dx}}\left( {{{{\log }}_{a}}u} \right)&=\frac{{{u}’}}{{u\left( {\ln \,a} \right)}}\\\frac{d}{{dx}}\left( {{{e}^{u}}} \right)&={{e}^{u}}{u}’\\\frac{d}{{dx}}\left( {{{a}^{u}}} \right)&=\left( {\ln \,a} \right){{a}^{u}}{u}’\end{align}\) \(\displaystyle \frac{d}{{dx}}\left[ {f{{{\left( x \right)}}^{{g\left( x \right)}}}} \right]:\text{take ln of each side}\) When we have a variable both in the base and the exponent, take ln of both sides to take derivative, and use implicit integration. Then substitute y function back in to get in terms of x.
Mean Value Theorem: If a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one number c in (a, b), where \(\displaystyle {f}’\left( c \right)=\frac{{f\left( a \right)f\left( b \right)}}{{ba}}\). (You may have to find the “c” by taking the derivative, setting to the slope you get on \(\left[ {a,\,b} \right]\) and solving for “c” in \(\left[ {a,\,b} \right]\).)  Intermediate Value Theorem (IVT): If a function f is continuous on a closed interval \(\left[ {a,\,b} \right]\), where \(f\left( a \right)\ne f\left( b \right)\), and m is any number between \(f\left( a \right)\) and \(f\left( b \right)\), there must be at least one number c in \(\left[ {a,\,b} \right]\) such that \(f\left( c \right)=m\).
Rolle’s Theorem: If a function is continuous on a closed interval and differentiable on the open interval (a, b), and \(f\left( a \right)=f\left( b \right)\) (the y’s on the endpoints are the same), then there is at least one number c in (a, b), where \({f}’\left( c \right)=0\).
Curve Sketching: Critical Point(s): point(s) where derivative is zero or undefined (critical points could also be endpoints) Local Minimum: \(\frac{{dy}}{{dx}}\) goes from negative to 0 (or undefined) to positive, or \(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}>0\) Local Maximum: \(\frac{{dy}}{{dx}}\) goes from positive to 0 (or undefined) to negative, or \(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}<0\) Point of Inflection (concavity changes): \(\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\) goes from positive to 0 (or undefined) to negative, or negative to 0 (or undefined) to positive. When looking at graphs:
Instantaneous vs. Average Rate of Change: Instantaneous rate of change (e.g., a velocity) between two points is the slope of the tangent line, which is the derivative at a point. Average rate of change over \(\left[ {a,\,b} \right]\) is the slope of the secant line, which is \(\frac{{f\left( b \right)f\left( a \right)}}{{ba}}\).  
Equation of Tangent Line at a Point:
Point of Horizontal (Vertical) Tangent Line:
Local (Tangent Line) Linearization:
Optimization:
Implicit Differentiation:
Related Rates Hints:
Error Approximation: To estimate the error (or % error) of a measurement,
Position, Velocity, and Acceleration:
Derivative of an Inverse Function: Let \(f\left( x \right)\) be a function that is differentiable on a certain interval. If \(f\left( x \right)\) has an inverse function \(g\left( x \right)\), and \(g\left( x \right)\) is differentiable for any value of x such that \({f}’\left( {g\left( x \right)} \right)\ne 0\), then \({g}’\left( x \right)=\frac{1}{{{f}’\left( {g\left( x \right)} \right)}}\). (If we want to find the derivative of the inverse of the function at a certain point “x”, we just find the “y” for the particular “x” in the original function, and use this value as the “x” in the derivative of this function. Then take the reciprocal of this number; this gives to get the derivative of the inverse of the original function at this point).
