MHT CET · Maths · Differentiation
If \(y=x \tan y\), then \(\frac{d y}{d x}=\)
- A \(\frac{\tan x}{x-y^2}\)
- B \(\frac{y}{x-x^2-y^2}\)
- C \(\frac{\tan x}{x-x^2-y^2}\)
- D \(\frac{\tan y}{y-x}\)
Answer & Solution
Correct Answer
(B) \(\frac{y}{x-x^2-y^2}\)
Step-by-step Solution
Detailed explanation
\(y=x \tan y \)
\( \therefore \frac{d y}{d x}=x \sec ^2 y \frac{d y}{d x}+\tan y \)
\( \therefore \quad\left(x \sec ^2 y-1\right) \frac{d y}{d x}=-\tan y \)
\( \therefore \frac{d y}{d x}=\frac{-\tan y}{x \sec ^2 y-1}=\frac{-x \tan y}{x^2 \sec ^2 y-x} \)
\( =\frac{-x \tan y}{x^2\left(1+\tan ^2 y\right)-x}=\frac{-x \tan y}{x^2+x^2 \tan ^2 y-x} \)
\( \therefore \frac{d y}{d x}=\frac{-y}{x^2+y^2-x}=\frac{y}{x-x^2-y^2}\) \(\cdots[\because y=x \tan y, \text { given }]\)
\( \therefore \frac{d y}{d x}=x \sec ^2 y \frac{d y}{d x}+\tan y \)
\( \therefore \quad\left(x \sec ^2 y-1\right) \frac{d y}{d x}=-\tan y \)
\( \therefore \frac{d y}{d x}=\frac{-\tan y}{x \sec ^2 y-1}=\frac{-x \tan y}{x^2 \sec ^2 y-x} \)
\( =\frac{-x \tan y}{x^2\left(1+\tan ^2 y\right)-x}=\frac{-x \tan y}{x^2+x^2 \tan ^2 y-x} \)
\( \therefore \frac{d y}{d x}=\frac{-y}{x^2+y^2-x}=\frac{y}{x-x^2-y^2}\) \(\cdots[\because y=x \tan y, \text { given }]\)
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