If y=x tan y,then frac{d y}{d x}= | vedclass

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

Question

(B) Given equation: $y = x 	an y$ Differentiating both sides with respect to $x$: $\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + 	an y$ Rearranging

Vedclass · Accepted Answer

$\frac{y}{x-x^2-y^2}$

Vedclass · Answer

(B) Given equation: $y = x 	an y$ Differentiating both sides with respect to $x$: $\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + 	an y$ Rearranging the terms to isolate $\frac{d y}{d x}$: $\frac{d y}{d x} - x \sec^2 y \frac{d y}{d x} = 	an y$ $\frac{d y}{d x} (1 - x \sec^2 y) = 	an y$ $\frac{d y}{d x} = \frac{	an y}{1 - x \sec^2 y}$ Multiply numerator and denominator by $x$: $\frac{d y}{d x} = \frac{x 	an y}{x - x^2 \sec^2 y}$ Since $y = x 	an y$,we have $	an y = \frac{y}{x}$: $\frac{d y}{d x} = \frac{y}{x - x^2 (1 + 	an^2 y)} = \frac{y}{x - x^2 - x^2 	an^2 y}$ Substituting $x^2 	an^2 y = y^2$: $\frac{d y}{d x} = \frac{y}{x - x^2 - y^2}$

If $y=x \tan y$,then $\frac{d y}{d x}=$

Similar Questions

If the slope of a tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ on it is $2$,then $a - b$ is

If $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}}$,then $\frac{dy}{dx}$ is equal to

If $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$, then $\frac{dy}{dx} = $

The two curves $x^{3}-3xy^{2}+2=0$ and $3x^{2}y-y^{3}=2$:

If $f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$, then $f'(3)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module