Question

If $\mathrm{y}={\mathrm{x}}^{\sin \mathrm{x}},$ then$\frac{\mathrm{dy}}{\mathrm{dx}}=$

• A. $\frac{{\mathrm{x}}^{\mathrm{sinx}}\left[\mathrm{x}\mathrm{cosx}\mathrm{logx}+\mathrm{sinx}\right]}{\mathrm{x}}$
• B. $\frac{{\mathrm{x}}^{\mathrm{sinx}}\left[\mathrm{xcosxlogx}+\cos \mathrm{x}\right]}{\mathrm{x}}$
• C. $\frac{\mathrm{y}\left[\mathrm{xsinxlogx}+\mathrm{cosx}\right]}{\mathrm{x}}$
• D. None of these

Answer 1

The correct option is A

$\frac{{\mathrm{x}}^{\mathrm{sinx}}\left[\mathrm{x}\mathrm{cosx}\mathrm{logx}+\mathrm{sinx}\right]}{\mathrm{x}}$

Explanation for the correct option:

Finding the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

The given function is $\mathrm{y}={\mathrm{x}}^{\sin \mathrm{x}}$

Apply the $\log$ on both sides of the given equation

$\mathrm{logy}=\mathrm{sinx}\mathrm{logx}\left[\because {\mathrm{loga}}^b=\mathrm{bloga}\right]$

Differentiate both sides of the given equation

$$\begin{gathered} \begin{array}{l}\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{sinx}\times \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)+\mathrm{logx}\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin x\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left(}\mathbf{logx}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{ab}\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}\mathbf{a}\frac{\mathbf{db}}{\mathbf{dx}}\mathbf{+}\mathbf{b}\frac{\mathbf{da}}{\mathbf{dx}}\mathbf{]}\\ \frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{sinx}\times \frac{1}{\mathrm{x}}+\mathrm{logx}\cos x\mathbf{[}\mathbf{\because }\frac{\mathbf{d}\left(\mathrm{sinx}\right)}{\mathbf{dx}}\mathbf{=}\mathbf{cosx}\mathbf{]}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left(\frac{\mathrm{sinx}}{\mathrm{x}}+\mathrm{logx}\cos x\right)\\ ={\mathrm{x}}^{\mathrm{sinx}}\left(\frac{\mathrm{sinx}}{\mathrm{x}}+\mathrm{logxcos}x\right)\mathbf{[}\mathbf{Given}\mathbf{}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{sinx}}\mathbf{]}\end{array} \\ =\frac{{\mathrm{x}}^{\mathrm{sinx}}\left(\mathrm{sinx}+\mathrm{xlogxcosx}\right)}{\mathrm{x}} \end{gathered}$$

Hence, the correct option is A.