Correct Answer - Option 2 :
\(\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)
Concept:
\(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)
\(\rm \frac{d(\tan^{-1} x)}{dx}= \frac{1}{1+x^2}\)
Calculation:
Given: y = \(\rm \tan^{-1}\left[ \frac{8x}{1-15x^2} \right ]\)
\(\rm y=\rm \tan^{-1}\left[ \frac{5x+3x}{1-5x\cdot 3x} \right ]\)
As we know that, \(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)
So, \(\rm y=\rm \tan^{-1}\left[ \frac{5x+3x}{1-5x\cdot 3x} \right ]\)= tan-1 5x + tan-1 x
Differentiating with respect to x, we get
\(\rm \frac{dy}{dx}=\frac{d(\tan^{-1} 5x)}{dx}+\frac{d(\tan^{-1} 3x)}{dx}\)
\(= \rm \frac{5}{1+(5x)^2}+\frac{3}{1+(3x)^2}\)
\(=\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)
