Inverse Trigonometry
Differentiation of Inverse Trig
Grade Class 12
Question:
If $y = \tan^{-1}\dfrac{4x}{1+5x^2} + \tan^{-1}\dfrac{2+3x}{3-2x}$, find $\dfrac{dy}{dx} = \dfrac{\alpha}{1+25x^2}$. Find $\alpha$.
Step-by-Step Solution
Key Concept: Split: $\tan^{-1}(4x/(1+5x^2)) = \tan^{-1}(5x)-\tan^{-1}(x)$; $\tan^{-1}((2+3x)/(3-2x))=\tan^{-1}(2/3)+\tan^{-1}(x)$. Sum simplifies.
$y = \tan^{-1}5x - \tan^{-1}x + \tan^{-1}\frac{2}{3} + \tan^{-1}x = \tan^{-1}5x + \tan^{-1}\frac{2}{3}$. So $\frac{dy}{dx} = \frac{5}{1+25x^2}$, giving $\alpha=5$.
Correct Answer: 3