If y^{cos x}=x^{sin y}, then frac{d y}{d x}= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $y^{\cos x}=x^{\sin y}$ Taking logarithm on both sides: $\cos x \log y = \sin y \log x$ Differentiating both sides with respect to

Vedclass · Accepted Answer

$\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}$

Vedclass · Answer

(A) Given equation: $y^{\cos x}=x^{\sin y}$ Taking logarithm on both sides: $\cos x \log y = \sin y \log x$ Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}(\cos x \log y) = \frac{d}{dx}(\sin y \log x)$ $(\cos x) \cdot \frac{1}{y} \frac{dy}{dx} + (\log y) \cdot (-\sin x) = (\sin y) \cdot \frac{1}{x} + (\log x) \cdot (\cos y) \frac{dy}{dx}$ Rearranging the terms to group $\frac{dy}{dx}$: $\frac{dy}{dx} \left( \frac{\cos x}{y} - \cos y \log x \right) = \frac{\sin y}{x} + \sin x \log y$ $\frac{dy}{dx} \left( \frac{\cos x - y \cos y \log x}{y} \right) = \frac{\sin y + x \sin x \log y}{x}$ $\frac{dy}{dx} = \frac{y(\sin y + x \sin x \log y)}{x(\cos x - y \cos y \log x)}$

If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$

Similar Questions

For $x>1$,if $(2 x)^{2 y}=4 e^{2 x-2 y}$,then $(1+\log 2 x)^2 \frac{d y}{d x}$ is equal to

Find $\frac{dy}{dx}$ for the function $y^{x} = x^{y}$.

For the curve $C : (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$,the value of $3y^{\prime}-y^{3}y^{\prime\prime}$ at the point $(\alpha, \alpha)$,where $\alpha > 0$,on $C$ is equal to:

If $\sqrt{y + x} + \sqrt{y - x} = c$ and $\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2} - 1}$, then the value of $K$ is

If $\sin^2 x + 2\cos y + xy = 0$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module