Name: Derivative of Functions in Parametric Form
Uploaded: 2019-07-19
Duration: 8 min 37 s
Description: Learn everything you'll need to know about derivatives of functions in parametric forms with this quick and easy lesson that'll cover the concept comprehensi...

Question

Find

\frac{dy}{dx}

when

y = \sin (\log \sqrt{x}) + x^{cosx}

Accepted Answer

Given that, $y = \sin (\log \sqrt{x}) + x^{\cos x}$

Let $u = \sin (\log \sqrt{x}) and v = x^{\cos x}$

So, $y = u + v$

$\Rightarrow \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$

$\Rightarrow \frac{d u}{d x} = \cos (\log \sqrt{x}) \times \frac{d}{d x} (\log \sqrt{x})$

$\Rightarrow \frac{d u}{d x} = \cos (\log \sqrt{x}) \times \frac{1}{\sqrt{x}} \times \frac{1}{2 \sqrt{x}} = \frac{\cos (\log \sqrt{x})}{2 x}$

$\Rightarrow v = x^{\cos x}$

Taking logarithm on both sides,

$\Rightarrow \log v = \log (x^{\cos x})$

$\Rightarrow \log v = \cos x \cdot \log x$

$\Rightarrow \frac{1}{v} \frac{d v}{d x} = \cos x \times \frac{1}{x} + \log x \times (- \sin x)$

$\Rightarrow \frac{d v}{d x} = v [\frac{\cos x}{x} - \sin x \cdot \log x]$

$\Rightarrow \frac{d v}{d x} = x^{\cos x} [\frac{\cos x}{x} - \sin x \cdot \log x]$

$\Rightarrow \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{\cos (\log \sqrt{x})}{2 x} + x^{\cos x} [\frac{\cos x - x \cdot \sin x \cdot \log x}{x}]$

Find $\frac{dy}{dx}$ when $y = \sin (\log \sqrt{x}) + x^{cosx}$

Important Questions on Differentiation [Part-II]

Learn and Prepare for any exam you want !

Find dydx when y=sin(logx)+xcosx

Important Questions on Differentiation [Part-II]

Learn and Prepare for any exam you want !

Find $\frac{dy}{dx}$ when $y = \sin (\log \sqrt{x}) + x^{cosx}$