009A Sample Final 1, Problem 3 Detailed Solution

Find the derivatives of the following functions.

(a)   $f(x)=\ln {\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}$

(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 g(x)=3\sin (4x)+4\tan (\sqrt{1+x^3})}

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Background Information:
1. Chain Rule
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{d}{dx}(f(g(x)))=f'(g(x))g'(x)}
2. Quotient Rule
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{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}}
3. Trig Derivatives
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{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x}

Solution:

(a)

Step 1:
Using the Chain Rule, we have
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{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\cdot\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\cdot\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg).}\\ \end{array}}

Step 2:
Now, we need to calculate  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{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg).}
To do this, we use the Quotient Rule. So, we have
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{\frac{x^2+1}{x^2-1}\cdot\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\cdot\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\cdot\frac{4x}{(x^2+1)^2}\cdot}\\ &&\\ & = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\ &&\\ & = & \displaystyle{\frac{4x}{x^4-1}.}\\ \end{array}}

(b)

Step 1:
We need to use the Chain Rule. We have
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{3\cos(4x)\cdot (4x)'+4\sec^2(\sqrt{1+x^3})\cdot \frac{d}{dx}(\sqrt{1+x^3})}\\ &&\\ & = & \displaystyle{3\cos(4x)\cdot 4+4\sec^2(\sqrt{1+x^3})\cdot \frac{d}{dx}(\sqrt{1+x^3})}\\ &&\\ & = & \displaystyle{12\cos(4x)+4\sec^2(\sqrt{1+x^3})\cdot \frac{d}{dx}(\sqrt{1+x^3})}\\ \end{array}}

Step 2:
We need to calculate   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{d}{dx}(\sqrt{1+x^3}).}
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{12\cos(4x)+4\sec^2(\sqrt{1+x^3})\cdot\frac{d}{dx}\sqrt{1+x^3}}\\ &&\\ & = & \displaystyle{12\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}}\cdot \frac{d}{dx}(1+x^3)}\\ &&\\ & = & \displaystyle{12\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}} (3x^2)}\\ &&\\ & = & \displaystyle{12\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}.}\\ \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 f'(x)=\frac{4x}{x^4-1}}
(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 g'(x)=12\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}}

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