Find {eq}f'(x) {/eq} for the following function, then find {eq}f'(2),\ f'(0) {/eq}, and {eq}f'(-26) {/eq}. {eq}\displaystyle f(x) = \frac{81}{x-1} {/eq}

Question:

Find {eq}f'(x) {/eq} for the following function, then find {eq}f'(2),\ f'(0) {/eq}, and {eq}f'(-26) {/eq}.

{eq}\displaystyle f(x) = \frac{81}{x-1} {/eq}

Differentiation Rules

To differentiate a rational function with the numerator, a constant, we can use more efficiently Chain rule.

Quotient rule is a differentiation rule applied to rational function, given as

{eq}\displaystyle \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)= \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}, {/eq} where {eq}\displaystyle '=\frac{d}{dx}. {/eq}

If the function is a composition of functions, then we will use Chain Rule,

{eq}\displaystyle \frac{d}{dx}\left(f(g(x))\right)= \frac{df}{dg}(g(x))\cdot \frac{dg}{dx}(x). {/eq}

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To find the first derivative, {eq}\displaystyle f'(x) {/eq} for the function {eq}\displaystyle f(x) = \frac{81}{x-1}, {/eq}

we will apply chain...