If log (x+y)=sin (x+y),then frac{dy}{dx} is | vedclass

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

Question

(D) Given equation: $\log (x+y) = \sin (x+y)$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = \cos (x+

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) Given equation: $\log (x+y) = \sin (x+y)$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = \cos (x+y) \left(1 + \frac{dy}{dx}\right)$ Rearranging the terms: $\left(1 + \frac{dy}{dx}\right) \left(\frac{1}{x+y} - \cos (x+y)\right) = 0$ Since $\log (x+y) = \sin (x+y)$,the term $\frac{1}{x+y} - \cos (x+y)$ cannot be zero for all $x, y$ in the domain. Therefore,we must have: $1 + \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -1$

If $\log (x+y)=\sin (x+y)$,then $\frac{dy}{dx}$ is

Similar Questions

If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{d y}{d x}$ is equal to

The curve $x^4-2xy^2+y^2+3x-3y=0$ cuts the $X$-axis at $(0,0)$ at an angle of

If $x^{p}+y^{q}=(x+y)^{p+q}$,then $\frac{dy}{dx}$ is

If $x \cdot \log _{e}(\log _{e} x)-x^2+y^2=4$ and $y>0$,then $\frac{dy}{dx}$ at $x=e$ is

If $f(x) = (\cos x)(\cos 2x) \ldots (\cos nx)$, then $f^{\prime}(x) + \sum_{r=1}^n (r \tan rx) f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module