Question

If $y = \tan^{-1}(\frac{\log(\frac{e}{x^2})}{\log(ex^2)}) + \tan^{-1}(\frac{4 + 2\log x}{1 - 8\log x})$, then $\frac{dy}{dx}$ is

Answer 1

0

Answer 2

1. Write the given expression as:

   $A = \dfrac{\log\left(\frac{e}{x^2}\right)}{\log(e x^2)} = \dfrac{1 - 2\log x}{1 + 2\log x}$

   $B = \dfrac{4+2\log x}{1-8\log x}$

2. Notice that

   $\tan^{-1}\Big(\dfrac{1-2\log x}{1+2\log x}\Big) = \frac{\pi}{4} - \tan^{-1}(2\log x)$

   (since $\tan\left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan\theta}{1+\tan\theta}$).

3. Also, using the tangent addition formula,

   $\tan^{-1}\Big(\frac{4+2\log x}{1-8\log x}\Big) = \tan^{-1}(4) + \tan^{-1}(2\log x)$

   (because $\tan\big(\tan^{-1}(4)+\tan^{-1}(2\log x)\big) = \frac{4+2\log x}{1-8\log x}$).

4. Thus, the sum becomes:

   $y = \left(\frac{\pi}{4} - \tan^{-1}(2\log x)\right) + \left(\tan^{-1}(4) + \tan^{-1}(2\log x)\right) = \frac{\pi}{4} + \tan^{-1}(4)$.

5. Since $y$ is a constant, its derivative with respect to $x$ is $0$.