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