Differential Calculus Quick Study Guide

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)}}{{x-c}}$

Constant Rule:

$ \displaystyle \frac{d}{{dx}}\left( c \right)=0$

Power Rule:

$ \displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n-1}}}$

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 \frac{d}{{dx}}\left( {\sin x} \right)=\cos x\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cos x} \right)=-\sin x$

$ \displaystyle \frac{d}{{dx}}\left( {\tan x} \right)={{\sec }^{2}}x\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right)=-{{\csc }^{2}}x$

$ \displaystyle \frac{d}{{dx}}\left( {\sec x} \right)=\sec x\cdot \tan x$

$ \displaystyle \frac{d}{{dx}}\left( {\csc x} \right)=-\csc x\cdot \cot x$

(Trig functions starting with “$ c$” are negative. I also remember that there are always two sec‘s (csc‘s) and one tan (cot‘s) in the last four 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, and use implicit integration to take derivative. Then substitute the original $ y$ function back in to get the derivative all in terms of $ x$.

Mean Value Theorem:

If a function is continuous on a closed interval $ \left[ {a,b} \right]$ and differentiable on the open interval $ \left( {a,\,b} \right)$, then there is at least one number c in $ \left( {a,\,b} \right)$, where

 $ \displaystyle {f}’\left( c \right)=\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}$.

(Instantaneous Rate of Change = Average Rate of Change)

(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$”; this value must be in the $ \left( {a,b} \right)$ interval.)

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:

(Special case of Mean Value Theorem) If a function is continuous on a closed interval $ \left[ {a,\,b} \right]$, 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: $ \displaystyle \frac{{dy}}{{dx}}$ goes from negative to $ 0$ (or undefined) to positive, or $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}>0$

Local Maximum: $ \displaystyle \frac{{dy}}{{dx}}$ goes from positive to $ 0$ (or undefined) to negative, or $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}<0$

Point of Inflection (concavity changes): $ \displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ goes from positive to $ 0$ (or undefined) to negative, or negative to $ 0$ (or undefined) to positive.

Monotonic: always increasing and decreasing

When looking at graphs:

Instantaneous vs. Average Rate of Change:

Instantaneous rate of change (“IROC”) (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 (“AROC”) over $ \left[ {a,b} \right]$ is the slope of the secant line, which is $ \displaystyle \frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}$.

1. Take the derivative of the function; this is the slope of the tangent line. Plug the “$ x$”-value of the point given into the derivative to get the actual (numerical) slope “$ m$”.
2. Write the equation of the tangent line in the form $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, using the original point $ \left( {{{x}_{1}},{{y}_{1}}} \right)$. (You may need to get the “$ {{y}_{1}}$” value using the original function, if only “$ {{x}_{1}}$” value is given). Simplify, if needed.

Point of Horizontal (Vertical) Tangent Line:

1. Take the derivative, and set the numerator to $ 0$(horizontal tangent, derivative $ 0$) or denominator to $ 0$ (vertical tangent, derivative undefined).
2. Solve for $ x$, and then get $ y$ from plugging in $ x$ into the original function.

Local (Tangent Line) Linearization (using Differentials to Approximate a Number):

1. Find a “close” point that works with the original function. For example, for $ f\left( x \right)=\sqrt{x}$, to find $ \sqrt{{4.02}}$, use point $ \left( {4,2} \right)$.
2. Take the derivative of the function; this is the slope of the tangent line. Plug the “$ x$”-value of the point from above (“$ 4$”) into the derivative to get the actual (numerical) slope “$ m$”. (Example: $ \displaystyle {y}’=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}=\frac{1}{2}{{\left( 4 \right)}^{{-\frac{1}{2}}}}=\frac{1}{2}\cdot \frac{1}{{\sqrt{4}}}=\frac{1}{4}$.)
3. Write the equation of the tangent line in the form $ y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)$, using the “close” point found above, and the slope found in step 2; plug in the actual $ x$-value to get the actual $ y$-value. (Example: $ \displaystyle y-2=m\left( {x-4} \right);\,\,y-2=\frac{1}{4}\left( {x-4} \right);\,\,y=\frac{1}{4}\left( {4.02-4} \right)+2=2.005\approx \sqrt{{4.02}}$.)

Optimization:

1. Solve in terms of one variable, and take the derivative; set to $ 0$ to get the minimum or maximum.
2. In the case of a closed interval, check the endpoints of the interval to make sure they aren’t lower or higher than the minimum or maximum found.

Implicit Differentiation:

1. Differentiate both sides of equation with respect to $ x$. When are you differentiating variables other than $ x$ (such as “$ y$”), remember to multiply that term by $ \displaystyle \frac{{dy}}{{dx}}\text{ (}{y}’)$.
2. Move any term with $ {y}’$ in it to the left side of the equation and move all other terms to the right side (even if they have $ x$’s and $ y$’s in them).
3. Factor out the $ {y}’$ on the left side of the equation to isolate and solve for $ {y}’$. Example: $ \displaystyle y={{y}^{2}}+xy;\,\,{y}’=2y{y}’+\left( {x\cdot {y}’+y\cdot 1} \right);{y}’-2y{y}’-x{y}’=y;{y}’=\frac{y}{{1-2y-x}}$.

Related Rates Hints:

1. Draw a picture, and label any amounts that could be changing (variables), and not changing (constants).
2. Write down exactly what the problem gives you, and what you need to solve for. For example, you may write down “Find $ \displaystyle \frac{{dA}}{{dt}}$ when $ \text{radius}=6$”. The rates (things that are changing) have “$ dt$” (with respect to time). For rates, make sure the sign is correct (for example, “filling up” is positive a volume rate, “emptying out” is a negative volume rate).
3. Write an equation that relates all of the given information and variables. For example, this may be a Geometry formula, such as $ A=\pi {{r}^{2}}$.
4. Simplify by trying to put everything in as few variables as possible before differentiating (you may have to substitute some variables by solving in terms of other variables). Typically, when you have values that say “when” something happens, these are put in after differentiating.
5. Use implicit differentiation to differentiate with respect to time. Fill in the equation with what you know (for example, the “when”s). If you can’t fill in all the variables that you need, you may need to use an additional formula, given what you have.
6. Solve the equation for what the problem is asking for.
7. Types of problems: Geometric Shapes typically use area and volume, Shadow problems typically use similar triangles, Right Triangle problems (such as Ladder problems) typically use the Pythagorean Theorem, and Problems with Angles typically use Trigonometry.

Error Approximation:

To estimate the error (or % error) of a measurement:

1. Attack this like a related rates problem: write down what we know, what we need, and how we relate the variables, such as with a Geometry formula, like volume.
2. Use implicit differentiation to take the derivative of the function given; the error given will be one of the “$ dx$” parts of the derivative (for example, change in volume might be $ dV$).
3. Solve for the error of the measurement needed, such as $ ds$. To get % error, divide by the total amount of this measurement.

Position, Velocity, and Acceleration:

• If a particle is moving along a horizontal line, its position $ p\left( t \right)$ (for example, relative to the origin) is represented by a function, the derivative of this function is its (instantaneous) velocity $ v\left( t \right)$ (how fast it’s moving), and the derivative of its velocity is its acceleration $ a\left( t \right)$ (how fast its velocity is changing). Displacement is the difference between the final and initial positions of a particle, and can be positive or negative. Distance is how far the particle is from its original position, and is always positive. (Example: $ p\left( t \right)=-16{{t}^{2}}+96t\text{ ft};\,\,v\left( t \right)=-32t+96\,\,\text{ft/sec};\,\,a\left( t \right)=-32\,\,\text{ft/se}{{\text{c}}^{2}}$.)
• Position, velocity, acceleration and displacement are vectors. Distance is the absolute value of the displacement vector and speed is the absolute value of the velocity vector; these are scalars. Speed is increasing when velocity and acceleration have the same sign, and decreasing when they have different signs.
• If the derivative (velocity) is positive, the object is moving to the right (or up, if that’s how the coordinate system is defined). If the velocity is negative, it’s moving to the left (or down). If the velocity is $ 0$, the object is at rest. Thus, if the velocity changes signs, the object changes direction.
• To find the maximum or minimum velocity or accelerations, find where the derivative of these functions fails or is zero.
• Average Velocity is $ \displaystyle \frac{{\text{Ending Position }-\text{ Initial Position}}}{{\text{Total}\,\,\text{Time}}}=\frac{{\Delta x}}{{\Delta t}}$.

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 $ \displaystyle {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$“-value for the particular “$ x$“-value in the original function, and use this value as the “$ x$“-value 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).