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  4. If f(x)=(x-4/2√x), then f'(1) is

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Q. If $f\left(x\right)=\frac{x-4}{2\sqrt{x}}$, then $f'\left(1\right)$ is

Limits and Derivatives

Solution:

We have, $f\left(x\right)=\frac{x-4}{2\sqrt{x}}$
$\therefore f'\left(x\right)=\frac{1}{2}\left\{\frac{d}{dx}\left(\frac{x-4}{\sqrt{x}}\right)\right\}$
$=\frac{1}{2} \frac{\left\{\sqrt{x} \frac{d}{dx}\left(x-4\right)-\left(x-4\right) \frac{d}{dx}\sqrt{x}\right\}}{\left(\sqrt{x}\right)^{2}}$
$=\frac{1}{2}\left\{\left(\sqrt{x}\times 1\right)-\left(x-4\right)\times\frac{1}{2\sqrt{x}}\right\} \times \frac{1}{x}$
$=\frac{1}{2}\left(\sqrt{x}-\frac{x-4}{2\sqrt{x}}\right) \frac{1}{x}$
$\therefore f'\left(1\right)=\frac{1}{2}\left(1-\frac{\left(-3\right)}{2}\right) \times1=\frac{5}{4}$