AP EAMCET · Maths · Inverse Trigonometric Functions
If \(y=\operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}\), then \(\frac{d y}{d x}=\)
- A \(-\frac{1}{2 \sqrt{1-x^2}}\)
- B \(\frac{-1}{2 x \sqrt{1-x^2}}\)
- C \(\frac{2}{1+x^2}\)
- D \(\frac{1}{2 x \sqrt{1+x^2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{-1}{2 x \sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
\(y=\operatorname{Tanh}^{-1} u\) where \(u = \sqrt{\frac{1-x}{1+x}}\) \(\frac{dy}{dx} = \frac{1}{1-u^2} \frac{du}{dx}\) \(1-u^2 = 1-\frac{1-x}{1+x} = \frac{1+x-(1-x)}{1+x} = \frac{2x}{1+x}\)…
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