TS EAMCET · Maths · Differentiation
If \(y^{\cos x}=x^{\sin y}\), then \(\frac{d y}{d x}=\)
- A \(\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}\)
- B \(\frac{y(x \sin x \log x-\sin y)}{x(\cos x+y \log x \cos y)}\)
- C \(\frac{y(\sin y-x \log y)}{x(x-y \cos y(\log x))}\)
- D \(\frac{y(\sin y+x \log y)}{x(x+y \cos y(\log x))}\)
Answer & Solution
Correct Answer
(A) \(\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}\)
Step-by-step Solution
Detailed explanation
It is given that \(y^{\cos x}=x^{\sin y}\) On taking logrithm both sides, we get \(\cos x \log y=\sin y \log x\) On differentiating both side w.r.t. \(x\), we get \(\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y\) \(=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x}\)…
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