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Wize University Calculus 2 Textbook > Review: Derivatives

Derivative Rule

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Derivative Rules

Power Rule

ddx(a)=0\frac{d}{dx}\left(a\right)=0 ddx(xn)=nxn−1\frac{d}{dx}\left(x^n\right)=nx^{n-1}

Exponential

ddx(ax)=axln⁡a\frac{d}{dx}\left(a^x\right)=a^x\ln a ddx(ex)=ex\frac{d}{dx}\left(e^x\right)=e^x

Logarithmic

ddx(log⁡ax)=1xln⁡a\frac{d}{dx}\left(\log_ax\right)=\frac{1}{x\ln a} ddx(ln⁡x)=1x\frac{d}{dx}\left(\ln x\right)=\frac{1}{x}

Trig

ddx(sin⁡x)=cos⁡x\frac{d}{dx}\left(\sin x\right)=\cos x ddx(cos⁡x)=−sin⁡x\frac{d}{dx}\left(\cos x\right)=-\sin x ddx(tan⁡x)=sec⁡2x\frac{d}{dx}\left(\tan x\right)=\sec^2x

ddx(csc⁡x)=−csc⁡xcot⁡x\frac{d}{dx}\left(\csc x\right)=-\csc x\cot x ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x)=\sec x\tan x ddx(cot⁡x)=−csc⁡2x\frac{d}{dx}(\cot x)=-\csc^2x

Inverse Trig

ddx(arcsin⁡x)=11−x2\frac{d}{dx}\left(\arcsin x\right)=\frac{1}{\sqrt{1-x^2}} ddx(arccos⁡x)=−11−x2\frac{d}{dx}\left(\arccos x\right)=-\frac{1}{\sqrt{1-x^2}} ddx(arctan⁡x)=11+x2\frac{d}{dx}\left(\arctan x\right)=\frac{1}{1+x^2}

Two More Important Rules

ddx(f±g)=ddx(f)±ddx(g)\frac{d}{dx}\left(f\pm g\right)=\frac{d}{dx}\left(f\right)\pm\frac{d}{dx}\left(g\right)
ddx(c⋅f)=c⋅ddx(f)\frac{d}{dx}\left(c\cdot f\right)=c\cdot\frac{d}{dx}\left(f\right)


You can keep taking derivatives to find second derivatives, third derivatives, etc.

Wize Tip
Always try to simplify before finding derivatives.

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Example: Derivative Rules

Find the derivative of the following:
a. f(x)=x3×x−2x+e2−3log⁡3xf\left(x\right)=x^3\times\sqrt{x}-2x+e^2-3\log_3x.
We first simplify:
f(x)=x3×x12−2x+e2−3log⁡3xf\left(x\right)=x^3\times x^{\frac{1}{2}}-2x+e^2-3\log_3x
f(x)=x72−2x+e2−3log⁡3xf\left(x\right)=x^{\frac{7}{2}}-2x+e^2-3\log_3x

Now we find the derivative one term at a time:
f′(x)=72x52−2+0−3(1x⋅ln⁡3)f'\left(x\right)=\frac{7}{2}x^{\frac{5}{2}}-2+0-3\left(\frac{1}{x\cdot\ln3}\right)
f′(x)=72x52−2−3xln⁡3f'\left(x\right)=\frac{7}{2}x^{\frac{5}{2}}-2-\frac{3}{x\ln3}


b. f(x)=sin⁡x−3cos⁡x+2arcsin⁡xf\left(x\right)=\sin x-3\cos x+2\arcsin x
This is already simplified as much as possible, we find the derivative one term at a time.
f′(x)=cos⁡x−3(−sin⁡x)+2(11−x2)f'\left(x\right)=\cos x-3\left(-\sin x\right)+2\left(\frac{1}{\sqrt{1-x^2}}\right)
f′(x)=cos⁡x+3sin⁡x+21−x2f'\left(x\right)=\cos x+3\sin x+\frac{2}{\sqrt{1-x^2}}
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Product, Quotient, and Chain Rule

Product Rule

ddx(f(x)g(x))=f′(x)g(x)+g′(x)f(x)\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)

Quotient Rule

ddx(f(x)g(x))=f′(x)g(x)−g′(x)f(x)(g(x))2\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)}{\left(g\left(x\right)\right)^2}

Chain Rule

ddx(f(g(x)))=f′(g(x))×g′(x)\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f'\left(g\left(x\right)\right)\times g'\left(x\right)

Strategies for finding derivatives:

  1. Simplify as much as you can
  2. Combine like terms
  3. Rewrite trignxtrig^nx as (trig x)n\left(trig\ x\right)^n
  4. Check if product, quotient, or chain rule is/are needed
  5. Determine which function type (derivative rule) is/are needed
  6. Simplify your answer after finding the derivative
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Example: Product, Quotient, and Chain Rule

Find the derivative of the following functions:
a) f(x)=x2sin⁡xf\left(x\right)=x^2\sin x
Since we have a product of 2 terms, we need product rule.
f′(x)=(2x)(sin⁡x)+(cos⁡x)(x2)f'\left(x\right)=\left(2x\right)\left(\sin x\right)+\left(\cos x\right)\left(x^2\right)
f′(x)=2xsin⁡x+x2cos⁡xf'\left(x\right)=2x\sin x+x^2\cos x

b) f(x)=sin⁡(x3)f\left(x\right)=\sin\left(x^3\right)
Since we have a function wrapped inside another, we need chain rule.
Starting from the outside function, then working our way in:
f′(x)=cos⁡(x3)×(3x2)f'\left(x\right)=\cos\left(x^3\right)\times\left(3x^2\right)
f′(x)=3x2cos⁡x3f'\left(x\right)=3x^2\cos x^3

c) f(x)=sin⁡3xf\left(x\right)=\sin^3x
Rewriting this, we get f(x)=(sin⁡x)3f\left(x\right)=\left(\sin x\right)^3
Sine we have a function wrapped inside another, we need chain rule.
Starting from the outside function, then working our way in:
f′(x)=3(sin⁡x)2×(cos⁡x)f'\left(x\right)=3\left(\sin x\right)^2\times\left(\cos x\right)
f′(x)=3sin⁡2x cos⁡xf'\left(x\right)=3\sin^2x\ \cos x

d) f(x)=etan⁡xf\left(x\right)=e^{\sqrt{\tan x}}
Rewriting this, we get f(x)=e(tan⁡x)12f\left(x\right)=e^{\left(\tan x\right)^{\frac{1}{2}}}
Since we have a function wrapped inside another, wrapped inside another, we need chain rule.
Starting from the outermost function, then working our way in:
f′(x)=[e(tan⁡x)12]×[12(tan⁡x)−12]×[sec⁡2x]f'\left(x\right)=\left[e^{\left(\tan x\right)^{\frac{1}{2}}}\right]\times\left[\frac{1}{2}\left(\tan x\right)^{-\frac{1}{2}}\right]\times\left[\sec^2x\right]

Now we simplify:
f′(x)=etan⁡xsec⁡2x2tan⁡xf'\left(x\right)=\frac{e^{\sqrt{\tan x}}\sec^2x}{2\sqrt{\tan x}}

Practice: Finding a Derivative

(tan⁡−1(3sin⁡x))′∣x=π=\left(\tan^{-1}\left(3^{\sin x}\right)\right)'|_{x=\pi}=

Practice: Finding a Derivative

Given that f(x)=(2−arcsin⁡(x2))3f\left(x\right)=\left(2-\arcsin\left(x^2\right)\right)^3, find f′(0)f'\left(0\right).
If f(x)=sin⁡(2x)f\left(x\right)=\sin\left(2x\right), find f(21)(π2)f^{\left(21\right)}\left(\frac{\pi}{2}\right).
(i.e. find the 21st derivative at the point π2\frac{\pi}{2})

Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?

Given this table of values for f, g, f′, and g′f,\ g,\ f',\ \text{and}\ g' below, answer the following questions.
xf(x)f′(x)f′′(x)g(x)g′(x)00−1−523π2π10452−2−410π2−3\begin{array}{|c|c|c|c|c|c|} \hline x&f(x)&f'(x)&f''(x)&g(x)&g'(x)\\ \hline 0&0&-1&-5&2&3\\ \hline \frac{\pi}{2}&\pi&1&0&4&5\\ \hline 2&-2&-4&10&\frac{\pi}{2}&-3\\ \hline \end{array}
If p(x)=f(g(x))p\left(x\right)=f\left(g\left(x\right)\right), then p′(2)=p'\left(2\right)=

Extra Practice
Practice: Derivative Rules (Level 2)
Find the rate of the change of the function f(x)=5x−x32+13x\displaystyle f(x)=\frac{5}{\sqrt{x}}-\frac{\sqrt[3]{x}}{2}+\frac{1}{3x} at the point x=1x=1.
Differentiation Rules
Find ddx(sin⁡2x+π+cos⁡2x+57)\frac{d}{dx}(\sin^2{x}+\sqrt\pi+\cos^2{x}+5^7)
Differentiation Rules
Find ddx(sin⁡2x+π+cos⁡2x+57)\frac{d}{dx}(\sin^2{x}+\sqrt\pi+\cos^2{x}+5^7)
Differentiation Rules
Find ddx(sin⁡2x+π+cos⁡2x+57)\frac{d}{dx}(\sin^2{x}+\sqrt\pi+\cos^2{x}+5^7)
Differentiation Rules
Find ddx(sin⁡2x+π+cos⁡2x+57)\frac{d}{dx}(\sin^2{x}+\sqrt\pi+\cos^2{x}+5^7)
Practice: Product Rule*
Suppose g(x)g\left(x\right) is differentiable, f(x)=x2g(x)f\left(x\right)=x^2g\left(x\right). Given that f′(3)=30f'\left(3\right)=30, f′′(3)=19f''\left(3\right)=19, g(3)=2g\left(3\right)=2, what is g′(3)g'\left(3\right) and g′′(3)g''\left(3\right)?
Practice: Quotient Rule
Find the derivative of f(x)=x2+3x4/3+1x2+1\displaystyle f(x)=\frac{x^2+3x^{4/3}+1}{x^2+1}
Practice: Derivative Rule (Level 3)
Practice: Derivative Rule
Given f(t)=2t3(t+1t)t\displaystyle f(t)=\frac{2t^3\left(t+\frac{1}{t}\right)}{\sqrt{t}}, find f′(4)f'\left(4\right).
Differentiation Rules
Find ddx(sin⁡2x+π+cos⁡2x+57)\frac{d}{dx}(\sin^2{x}+\sqrt\pi+\cos^2{x}+5^7)
Practice: Multiple Chain Rules
Practice: Multiple Chain Rules
Find the derivative of f(x)=x+x4+x3+x2+1f\left(x\right)=x+\sqrt{x^4+\sqrt{x^3+\sqrt{x^2+1}}} at x=0x=0
Practice: Derivative Rules (Level 2)
Practice: Derivative Rules
Find the rate of the change of the function f(x)=5x−x32+13x\displaystyle f(x)=\frac{5}{\sqrt{x}}-\frac{\sqrt[3]{x}}{2}+\frac{1}{3x} at the point x=1x=1.