Chain Rule Mixed Practice

Practice Problems

 

  1. Differentiate

ddxecos(3x)\frac{d}{dx}\, e^{\cos(3x)}

  1. Differentiate

ddxsin(x2+1)\frac{d}{dx}\,\sin(x^2+1)

  1. Differentiate

ddx(cos(x3))2\frac{d}{dx}\,\big(\cos(x^3)\big)^2

  1. Differentiate

ddxln(5x2+1)\frac{d}{dx}\,\ln(5x^2+1)

  1. Differentiate

ddx1+3x4\frac{d}{dx}\,\sqrt{\,1+3x^4\,}

  1. Differentiate

ddx(2x+1)10\frac{d}{dx}\,(2x+1)^{10}

  1. Differentiate

ddxarctan ⁣(1x2)(assume x<1)\frac{d}{dx}\,\arctan\!\big(\sqrt{\,1-x^2\,}\big) \qquad(\text{assume }|x|<1)

  1. Differentiate

ddxloga(3x+5)(a>0, a1)\frac{d}{dx}\,\log_a(3x+5) \quad (a>0,\ a\neq 1)

  1. Differentiate

ddxex2sinx

\frac{d}{dx}\,e^{\,x^2\sin x}

  1. Differentiate (away from points where sin(2x)=0\sin(2x)=0)

ddxsin(2x)\frac{d}{dx}\,\big|\sin(2x)\big|

 

  1. If ff is differentiable and a,ba,b are constants, find

ddxf(ax2+b)\frac{d}{dx}\,f(ax^2+b)

 

Solutions

1)

ddxecos(3x)=ecos(3x)ddx(cos(3x))=ecos(3x)(sin(3x)3)=3sin(3x)ecos(3x).\frac{d}{dx}\, e^{\cos(3x)} = e^{\cos(3x)}\cdot \frac{d}{dx}\big(\cos(3x)\big) = e^{\cos(3x)}\cdot\big(-\sin(3x)\cdot 3\big) = -3\sin(3x)\,e^{\cos(3x)}.

 

2)

ddxsin(x2+1)=cos(x2+1)ddx(x2+1)=2xcos(x2+1).\frac{d}{dx}\,\sin(x^2+1) = \cos(x^2+1)\cdot \frac{d}{dx}(x^2+1) = 2x\,\cos(x^2+1).

 

3) Write u=cos(x3)u=\cos(x^3). Then ddx(u2)=2uu\frac{d}{dx}(u^2)=2u\,u’.

ddx(cos(x3))2=2cos(x3)(sin(x3)3x2)=6x2sin(x3)cos(x3).\frac{d}{dx}\,\big(\cos(x^3)\big)^2 = 2\cos(x^3)\cdot\big(-\sin(x^3)\cdot 3x^2\big) = -6x^2\sin(x^3)\cos(x^3).

 

4)

ddxln(5x2+1)=15x2+1(10x)=10x5x2+1.\frac{d}{dx}\,\ln(5x^2+1) = \frac{1}{5x^2+1}\cdot (10x) = \frac{10x}{5x^2+1}.

 

5) Write 1+3x4=(1+3x4)1/2\sqrt{\,1+3x^4\,}=(1+3x^4)^{1/2}.

ddx(1+3x4)1/2=12(1+3x4)1/212x3=6x31+3x4.\frac{d}{dx}\,(1+3x^4)^{1/2} = \tfrac12(1+3x^4)^{-1/2}\cdot 12x^3 = \frac{6x^3}{\sqrt{\,1+3x^4\,}}.

 

6) Power rule with inner derivative:

ddx(2x+1)10=10(2x+1)92=20(2x+1)9.\frac{d}{dx}\,(2x+1)^{10} = 10(2x+1)^9\cdot 2 = 20(2x+1)^9.

 

7) Let u=1x2u=\sqrt{1-x^2}. Then ddxarctan(u)=u1+u2\dfrac{d}{dx}\arctan(u)=\dfrac{u’}{1+u^2}.
Here u=x1x2u’=\dfrac{-x}{\sqrt{1-x^2}} and 1+u2=1+(1x2)=2x21+u^2=1+(1-x^2)=2-x^2.

ddxarctan ⁣(1x2)=x/1x22x2=x(2x2)1x2(x<1).\frac{d}{dx}\,\arctan\!\big(\sqrt{1-x^2}\big) = \frac{-x/\sqrt{1-x^2}}{\,2-x^2\,} = \frac{-x}{(2-x^2)\sqrt{1-x^2}} \quad(|x|<1).

 

8) ddxloga(3x+5)=1(3x+5)lna3\dfrac{d}{dx}\log_a(3x+5)=\dfrac{1}{(3x+5)\ln a}\cdot 3.

ddxloga(3x+5)=3(3x+5)lna.\frac{d}{dx}\,\log_a(3x+5)=\frac{3}{(3x+5)\ln a}.

 

9) Let u=x2sinxu=x^2\sin x. Then ddxeu=euu\dfrac{d}{dx}e^{u}=e^{u}u’.
Product rule for uu: u=2xsinx+x2cosxu’=2x\sin x+x^2\cos x.

ddxex2sinx=ex2sinx(2xsinx+x2cosx).\frac{d}{dx}\,e^{x^2\sin x} = e^{x^2\sin x}\big(2x\sin x+x^2\cos x\big).

 

10) For u(x)=sin(2x)u(x)=\sin(2x) and u(x)0u(x)\neq 0, ddxu=uuu\dfrac{d}{dx}|u|=\dfrac{u}{|u|}\,u’.
Here u=2cos(2x)u’ = 2\cos(2x).

ddxsin(2x)=sin(2x)sin(2x)2cos(2x),undefined where sin(2x)=0.\frac{d}{dx}\,|\sin(2x)| = \frac{\sin(2x)}{|\sin(2x)|}\cdot 2\cos(2x), \quad \text{undefined where }\sin(2x)=0.

 

11) Let u=ax2+bu=ax^2+b. Then

ddxf(ax2+b)=f(u)u=f(ax2+b)(2ax).\frac{d}{dx}\,f(ax^2+b)=f'(u)\cdot u’=f'(ax^2+b)\cdot (2ax).

 

When in doubt, name the inside u(x)u(\,x\,) and differentiate in two beats: outer derivative at uu, then multiply by uu’. That simple habit saves a lot of mistakes.

 

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