009A Sample Midterm 2, Problem 5

Find the derivatives of the following functions. Do not simplify.

a) $f(x)=\tan ^{3}(7x^{2}+5)$

b) $g(x)=\sin(\cos(e^{x}))$

c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} }

Foundations:
1. Chain Rule
2. Derivatives of trig/ln
3. Quotient Rule

Solution:

(a)

Step 1:
First, we use the Chain Rule to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.}

Step 2:
Now, we use the Chain Rule again to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{3\tan^2(7x^2+5)(\tan(7x^2+5))'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(7x^2+5)'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x).} \end{array}}

(b)

Step 1:
First, we use the Chain Rule to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=\cos(\cos(e^x))(\cos(e^x))'.}

Step 2:
Now, we use the Chain Rule again to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{\cos(\cos(e^x))(\cos(e^x))'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x)'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x).} \end{array}}

(c)

Step 1:
First, we use the Quotient Rule to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}.}

Step 2:
Now, we use the Chain Rule to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(5x^2+7x)'-(5x^2+7x)^2\frac{1}{x^2+1}(x^2+1)'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}.} \end{array}}

Final Answer:
(a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)}
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\cos(e^x))(-\sin(e^x))(e^x)}
(c)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}}

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