Question

If y = $x^x$ + $(sinx{)}^{cotx}$ . find $\frac{dy}{dx}$

Answer 1

$\frac{dy}{dx}=\frac{d(x{)}^x}{dx}+\frac{(sinx{)}^{cotx}}{dx}$

$\Rightarrow$ Let $z=x^x$

Take log on both sides

$\Rightarrow$$logz=xlogx$

Now differentiate above equation with respect to $x$

$\Rightarrow$$\frac{1}{z}\times \frac{dz}{dx}$ =$1+logx$

$\Rightarrow$$\frac{dz}{dx}=x^x(1+logx)$

$\Rightarrow$Let $t=(sinx{)}^{cotx}$

Take log on both sides

$\Rightarrow$$logt=cotx\times log\left(sinx\right)$

Now differentiate with respect to $x$

$\Rightarrow$$\frac{1}{t}\times \frac{dt}{dx}=-cose{c}^{2}x\times log\left(sinx\right)+cotx\times \frac{1}{sinx}\times cosx$

$\Rightarrow$$\frac{dt}{dx}$= $(sinx{)}^{cotx}\times (-cose{c}^{2}x\times log\left(sinx\right)+co{t}^{2}x)$

$\Rightarrow$$\frac{dy}{dx}=\frac{dt}{dx}+\frac{dz}{dx}$

Put the values of $\frac{dz}{dx}and\frac{dt}{dx}$

$\frac{dy}{dx}=x^x(1+logx)$$+$$(sinx{)}^{cotx}\times (-cose{c}^{2}x\times log\left(sinx\right)+co{t}^{2}x)$