Find \(\rm \frac{dy}{dx}\), if y = \(\rm \tan^{-1}\left[ \frac{8x}{1

1.	Find \(\rm \frac{dy}{dx}\), if y = \(\rm \tan^{-1}\left[ \frac{8x}{1-15x^2} \right ]\)1. \(\rm \frac{5}{1+25x^2}-\frac{3}{1+9x^2}\)2. \(\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)3. \(\rm \frac{8}{1+25x^2}\)4. None of these
Answer» 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 \(\rm \frac{dy}{dx}=\frac{d(\tan^{-1} 5x)}{dx}+\frac{d(\tan^{-1} 3x)}{dx}\) \(= \rm \frac{5}{1+(5x)^2}+\frac{3}{1+(3x)^2}\) \(=\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)

Find \(\rm \frac{dy}{dx}\), if y = \(\rm \tan^{-1}\left[ \frac{8x}{1-15x^2} \right ]\)1. \(\rm \frac{5}{1+25x^2}-\frac{3}{1+9x^2}\)2. \(\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)3. \(\rm \frac{8}{1+25x^2}\)4. None of these

Answer» 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

\(\rm \frac{dy}{dx}=\frac{d(\tan^{-1} 5x)}{dx}+\frac{d(\tan^{-1} 3x)}{dx}\)

\(= \rm \frac{5}{1+(5x)^2}+\frac{3}{1+(3x)^2}\)

\(=\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)