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In this question, you will prove the quotient rule for derivatives using the product and chain rules.

Let $h$ and $g$ be differentiable functions, with $g(x) > 0$ for all $x$, and let $f(x)=x^{-1}$.

First, calculate the derivative, $f'(x)$. Then, note that the quotient $\frac{h(x)}{g(x)}$ can be written as the composite, $h(x)[(f \circ g)(x)] = h(x)f(g(x)).$

Then, use the product and chain rules to derive $\frac{d}{dx}\frac{h}{g} = \frac{h'g - hg'}{g^2}$.

Thomas
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castleminer
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1 Answers1

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Hint(s): $$ f'(x) = -x^{-2} $$ You have using the product rule $$ \frac{d}{dx} \frac{h(x)}{g(x)} = \frac{d}{dx} h(x)f(g(x)) = h'(x)f(g(x)) + h(x)\frac{d}{dx}f(g(x)) $$ Now you just have to apply the chain rule and the simplify.

Thomas
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