Find the derivatives of the following functions.
a)
b)
| Foundations:
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| Review chain rule, quotient rule, and derivatives of trig functions
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Solution:
(a)
| Step 1:
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| Using the chain rule, we have
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- 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)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ \end{array}}
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| Step 2:
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| 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 \bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}
.
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| To do this, we use the Chain Rule. So, we have
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- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {4x}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {4x}{(x^{2}-1)(x^{2}+1)}}\\&&\\&=&\displaystyle {\frac {4x}{x^{4}-1}}\\\end{array}}}
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(b)
| Final Answer:
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(a)
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| (b)
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