Continuity and Differentiability — Class 12 Maths Solution

$y = {x^x} - {2^{\sin x}}$
$\Rightarrow$ $y = u - v,$ where $u = {x^x}$

and $v = {2^{\sin x}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} - \cfrac{{dv}}{{dx}}$ …(i)
Now, $u = {x^x}$

By taking log on both sides , we get

log u $=$ x log x

Differentiating w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{u}\cfrac{{du}}{{dx}} = x\cfrac{1}{x} + \log x$

$\Rightarrow$ $\cfrac{{du}}{{dx}} = {x^x}[1 + \log x]$ ….(ii)
and $v = {2^{\sin x}}$

By taking log on both sides , we get
$\Rightarrow$ $\log v = \sin x\log 2$

Differentiating w.r.t.x, we get

$\Rightarrow$ $\cfrac{1}{v}\cfrac{{dv}}{{dx}} = \log 2(\cos x)$
$\Rightarrow$ $\cfrac{{dv}}{{dx}} = {2^{\sin x}}(\cos x \cdot \log 2)$ ….(iii)

From (i), (ii) and (iii), we get

$\cfrac{{dy}}{{dx}} = {x^x}[1 + \log x] - {2^{\sin x}}(\cos x \cdot \log 2)$