If z = y + f(v),where v = frac{x}{y},then v f | vedclass

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

Question

(B) Given $z = y + f(v)$ and $v = \frac{x}{y}$.Substituting $v$,we have $z = y + f\left(\frac{x}{y}\right)$.Now,differentiate $z$ partially with respe

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $z = y + f(v)$ and $v = \frac{x}{y}$.Substituting $v$,we have $z = y + f\left(\frac{x}{y}\right)$.Now,differentiate $z$ partially with respect to $x$:$\frac{\partial z}{\partial x} = f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}$.Next,differentiate $z$ partially with respect to $y$ using the chain rule:$\frac{\partial z}{\partial y} = 1 + f'\left(\frac{x}{y}\right) \cdot \left(-\frac{x}{y^2}\right) = 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.Now,calculate the expression $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$:$v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \left(\frac{x}{y}\right) \left[f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}\right] + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.$= \frac{x}{y^2} f'\left(\frac{x}{y}\right) + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.$= 1$.

If $z = y + f(v)$,where $v = \frac{x}{y}$,then $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$ is

Similar Questions

If $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$,then $\sum \frac{\partial^2 u}{\partial x^2} = $

If $z = \sin^{-1}\left( \frac{x+y}{\sqrt{x} + \sqrt{y}} \right)$,then $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y}$ is equal to

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

If $u = u(x, y) = \sin(y + ax) - (y + ax)^2$, then which of the following is true?

If $u = x^2 + y^2$ and $x = s + 3t, y = 2s - t,$ then $\frac{d^2u}{ds^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module