Question

Given $\mathbf{x}\mathbf{=}\mathbf{In}\left(u^2,v^2\right)\mathbf{,}\mathbf{u}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$and $\mathbf{v}\mathbf{=}\mathbf{cos}\mathbf{}\mathbf{t}\mathbf{.}\mathbf{find}\frac{\mathbf{dx}}{\mathbf{dt}}$.

Short answer

The value of$\frac{dx}{dt}$is $\frac{4ut+2v\sin t}{u^2-v^2}$.

Step-by-step solution

Explanation of solution

The provided expressions are $\mathrm{x}=\mathrm{In}\left({\mathrm{u}}^2-{\mathrm{v}}^2\right),\mathrm{u}={\mathrm{t}}^2\mathrm{and}\mathrm{v}=\mathrm{cost}$.

Chain Rule

The number of functions in the composition affects how many differentiation steps are required when using the chain rule to get the derivative of composite functions.

Calculation

Consider the below equation to find $dx$.

$dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$

Find$\frac{\partial x}{\partial u}$.

$$\begin{gathered} \frac{\partial x}{\partial u}=\frac{1}{u^2-v^2}2u \\ =\frac{2u}{u^2-v^2} \end{gathered}$$

Find$\frac{\partial x}{\partial v}$.

$$\begin{gathered} \frac{\partial x}{\partial v}=\frac{1}{u^2-v^2}\left(-2v\right) \\ =\frac{-2v}{u^2-v^2} \end{gathered}$$

Find dx.

$dx=\frac{-2v}{u^2-v^2}du-\left(\frac{2u}{u^2-v^2}\right)dv$

Use the value of $du=2dt$and $dv=-\sin tdt$also differentiate above equation with respect to t.

$$\begin{gathered} \frac{dx}{dt}=\frac{2v}{u^2-v^2}\left(2t\right)-\left(\frac{2v}{u^2-v^2}\right)\left(-\sin t\right) \\ =\frac{4ut+2v\sin t}{u^2-v^2} \end{gathered}$$

Hence the value of$\frac{dx}{dt}$ is $\frac{4ut+2v\sin t}{u^2-v^2}$.