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 $\frac{d y}{d x}$ , when $y = x^{\sin x} + {(\sin x)}^{x}$

Accepted Answer

Let $y = x^{\sin x} + {(\sin x)}^{x}$

Also, let $u = x^{\sin x}$ and $v = {(\sin x)}^{x}$

$∴ y = u + v$

$\Rightarrow \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x} . . . (i)$

Now, $u = x^{\sin x}$

Taking $\log$ on both sides,

$\Rightarrow \log u = \log (x^{\sin x})$

$\Rightarrow \log u = \sin x \log x$

Differentiating both sides with respect to $x$ ,

$\frac{1}{u} \frac{d u}{d x} = \log x \frac{d}{d x} (\sin x) + \sin x \frac{d}{d x} (\log x)$

$\Rightarrow \frac{d u}{d x} = u [\cos x \log x + \sin x \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x} = x^{\sin x} [\cos x \log x + \frac{\sin x}{x}] . . . (ii)$

Again, $v = {(\sin x)}^{x}$

Taking $\log$ on both sides,

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

$\Rightarrow \log v = x \log (\sin x)$

Differentiating both sides with respect to $x$ ,

$\frac{1}{v} \frac{d v}{d x} = \log (\sin x) \frac{d}{d x} (x) + x \frac{d}{d x} [\log (\sin x)]$

$\Rightarrow \frac{d v}{d x} = v [\log (\sin x) + x \frac{1}{\sin x} \frac{d}{d x} (\sin x)]$

$\Rightarrow \frac{d v}{d x} = {(\sin x)}^{x} [logsin x + \frac{x}{\sin x} \cos x]$

$\Rightarrow \frac{d v}{d x} = {(\sin x)}^{x} [logsin x + x \cot x] . . . . (iii)$

From $(i), (ii)$ and $(iii)$ , we obtain

$\frac{d y}{d x} = x^{\sin x} (\cos x \log x + \frac{\sin x}{x}) + {(\sin x)}^{x} [logsin x + x \cot x]$

Find $\frac{d y}{d x}$ , when $y = x^{\sin x} + {(\sin x)}^{x}$

Important Questions on Differentiation

Learn and Prepare for any exam you want !

Find dydx, when y=xsinx+sinxx

Important Questions on Differentiation

Learn and Prepare for any exam you want !

Find $\frac{d y}{d x}$ , when $y = x^{\sin x} + {(\sin x)}^{x}$