Question

If $f\left(x,y\right)=\frac{\cos \left(x-4y\right)}{\cos \left(x+4y\right)}$, then $\frac{\partial f}{\partial x}$ at $\left(0,\frac{\mathrm{\pi }}{4}\right)$ is equal to

• A. $-1$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Step 1: Solve for the required partial derivative of the given function:

The given function is $f\left(x,y\right)=\frac{\cos \left(x-4y\right)}{\cos \left(x+4y\right)}$.

Partially differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{\partial }{\partial x}\left[f\left(x,y\right)\right]=\frac{\partial }{\partial x}\left[\frac{\cos \left(x-4y\right)}{\cos \left(x+4y\right)}\right] \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{\cos \left(x+4y\right)·\frac{\partial }{\partial x}\left[\cos \left(x-4y\right)\right]-\cos \left(x-4y\right)·\frac{\partial }{\partial x}\left[\cos \left(x+4y\right)\right]}{{\cos }^2\left(x+4y\right)}\mathbf{}\left[\because \frac{\partial }{\partial x}\left(\frac{u}{v}\right)=\frac{v·\frac{\partial u}{\partial x}-u·\frac{\partial v}{\partial x}}{v^2}\right] \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{\cos \left(x+4y\right)·\left[-\sin \left(x-4y\right)\right]\frac{\partial }{\partial x}\left(x-4y\right)-\cos \left(x-4y\right)·\left[-\sin \left(x+4y\right)\right]\frac{\partial }{\partial x}\left(x+4y\right)}{{\cos }^2\left(x+4y\right)}\left[\because \frac{\partial }{\partial x}f\left[g\left(x\right)\right]=f\text{'}\left[g\left(x\right)\right]·g\text{'}\left(x\right)\right] \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{-\cos \left(x+4y\right)·\sin \left(x-4y\right)·\left(1\right)+\cos \left(x-4y\right)·\sin \left(x+4y\right)·\left(1\right)}{{\cos }^2\left(x+4y\right)} \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{-\cos \left(x+4y\right)·\sin \left(x-4y\right)+\cos \left(x-4y\right)·\sin \left(x+4y\right)}{{\cos }^2\left(x+4y\right)} \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{\sin \left(x+4y\right)·\cos \left(x-4y\right)-\sin \left(x-4y\right)·\cos \left(x+4y\right)}{{\cos }^2\left(x+4y\right)} \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{\sin \left[\left(x+4y\right)-\left(x-4y\right)\right]}{{\cos }^2\left(x+4y\right)}\left[\because sin\left(A-B\right)=sin\left(A\right)·cos\left(B\right)-sin\left(B\right)·cos\left(A\right)\right] \\ \Rightarrow \frac{\partial f}{\partial x}=\frac{\sin \left(8y\right)}{{\cos }^2\left(x+4y\right)} \end{gathered}$$

Step 2: Solve for the required value:

The value of the given function at the point $\left(0,\frac{\mathrm{\pi }}{4}\right)$ can be given by,

$\frac{\partial f}{\partial x}\left(0,\frac{\mathrm{\pi }}{4}\right)=\frac{\sin \left(8\times {\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{{\cos }^2\left[0+\left(\frac{\mathrm{\pi }}{4}\times 4\right)\right]}$

$$\begin{gathered} =\frac{\sin \left({\displaystyle 2\mathrm{\pi }}\right)}{{\cos }^2\left(\mathrm{\pi }\right)} \\ =\frac{0}{{\left(-1\right)}^2} \\ =0 \end{gathered}$$

Therefore, $\frac{\partial f}{\partial x}$ at $\left(0,\frac{\mathrm{\pi }}{4}\right)$ is equal to $0$.

Hence, option(B) is the correct option.