# Assume that g(x)= \frac{f(x)}{x} and that f(x) is everywhere differentiable. Assuming that f(a)=1...

## Question:

Assume that {eq}g(x)= \frac{f(x)}{x}{/eq} and that {eq}f(x){/eq} is everywhere differentiable. Assuming that {eq}f(a)=1{/eq} and that {eq}f'(1)= \frac{1}{2}{/eq}, find an expression for the derivative of {eq}g(x){/eq} with respect to {eq}x{/eq} at {eq}x=1{/eq}, i.e., {eq}g'(x)= \frac{[d(g(x))]}{(dx)}{/eq} at {eq}x=1{/eq}.

## Differentiable Unknown Function:

If we have an unknown function such as {eq}h(t) {/eq} and it is differentiable so the derivative of this function with respect to the variable {eq}t {/eq} is equal to {eq}f'(t) {/eq}.

By the above rule of derivatives and quotient rule of derivatives in the given quotient expression, we'll get the required derivative function.

- {eq}\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{h(t)}{i(t)} \right )=\frac{i(t)h'(t)-h(t)i'(t)}{(i(t))^2} {/eq}

## Answer and Explanation:

The given quotient expression with the unknown function is:

{eq}g(x)=\displaystyle \frac{f(x)}{x} {/eq}

The given values of differentiable function {eq}f(x) {/eq} and its derivative at {eq}x=1 {/eq} are:

{eq}f(1)=1\\ f'(1)= \frac{1}{2} {/eq}

The differentaition of the function {eq}g(x) {/eq} with respect to {eq}x {/eq} by the quotient rule of derivatives is:

{eq}\begin{align*} \displaystyle \frac{[d(g(x))]}{(dx)}&=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f(x)}{x} \right )\\ g'(x)&=\frac{x\frac{\mathrm{d} f(x)}{\mathrm{d} x}-f(x)\frac{\mathrm{d} (x)}{\mathrm{d} x}}{x^2}\\ &=\frac{xf'(x)-f(x)(1)}{x^2}\\ &=\frac{xf'(x)-f(x)}{x^2}\\ \end{align*} {/eq}

Substituting {eq}1 {/eq} for {eq}x {/eq} in the above derivative function, we get:

{eq}\begin{align*} \displaystyle g'(1)&=\frac{1f'(1)-f(1)}{1^2}\\ &=\frac{f'(1)-f(1)}{1}\\ &=f'(1)-f(1) \end{align*} {/eq}

Plug the integer values of functions {eq}f(1) {/eq} and {eq}f'(1) {/eq} in the above expression and simplify it.

{eq}\begin{align*} \displaystyle g'(1)&=\frac{1}{2}-1\\ &=\displaystyle \frac{1-2}{2}\\ &=\displaystyle -\frac{1}{2}\\ \end{align*} {/eq}

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