Question

If$\mathit{w}\mathbf{=}{\left(rcos\theta \right)}^{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}$, find$\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{\theta }}$.

Short answer

The value of$\frac{\partial w}{\partial \theta }\text{is}\left[x\text{In}x-\frac{y^2}{x}\right]x^y$ .

Step-by-step solution

Given Information

Given that, $w={\left(r\cos \theta \right)}^{r\sin \theta }$.

Concept

To find out the value of $\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{\theta }}$ take logarithm on both the sides.

Finding the value of ∂w∂θ

Take logarithm on both the sides

$\log w=\log \left({\left(r\cos \theta \right)}^{\left(r\sin \theta \right)}\right)$

This can also be written as$\log w=\left(r\sin \theta \right)\log \left(r\cos \theta \right)$

Now differentiating the above function with respect to

$$\begin{gathered} \frac{1}{w}\frac{\partial w}{\partial \theta }=\left(r\sin \theta \right)\frac{-r\sin \theta }{r\cos \theta }+\log \left(r\cos \theta \right)r\cos \theta \\ \frac{\partial w}{\partial \theta }=\left[\log \left(r\cos \theta \right).r\cos \theta -\frac{r^2{\sin }^2\theta }{r\cos \theta }\right]{\left(r\cos \theta \right)}^{\left(r\sin \theta \right)} \\ \frac{\partial w}{\partial \theta }=\left[x\text{In}x-\frac{y^2}{x}\right]x^y \end{gathered}$$

Where, $x=r\cos \theta ,y=r\sin \theta$.

Hence$\frac{\partial w}{\partial \theta }=\left[x\text{In}x-\frac{y^2}{x}\right]x^y$ .