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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If z=(y)/(x) [ (x)/(y)+ (1+(y)/(x))], then x (∂ z)/(∂ x)is equal to

If z=yxz=\frac{y}{x}\left[\sin \frac{x}{y}+\cos\lleft(1+\frac{y}{x}\right)],thenthenx \frac{\partial z}{\partial x}is equal to

Options

  1. Ay∂z∂yy \frac{\partial z}{\partial y}
  2. B−y∂z∂y-y \frac{\partial z}{\partial y}
  3. C2y∂z∂y2 y \frac{\partial z}{\partial y}
  4. D2y∂z∂x2 y \frac{\partial z}{\partial x}

Correct answer

B. −y∂z∂y-y \frac{\partial z}{\partial y}

Step-by-step solution

We have, z=\frac{y}{x}\left[\sin \frac{x}{y}+\cos \lleft(1+\frac{y}{x}\right)] ∂z∂x=yx[cos⁡xy⋅1y−sin⁡\lleft(1+yx)(−yx2)]+(−yx2)[sin⁡xy+cos⁡\lleft(1+yx)]x∂z∂xx=y[1ycos⁡xy+yx2sin⁡\lleft(1+yx)]−yx[sin⁡xy+cos⁡\lleft(1+yx)]⇒x∂z∂x=cos⁡xy+y2x2sin⁡\lleft(1+yx)−z\begin{array}{r} \frac{\partial z}{\partial x}=\frac{y}{x} \left[\cos \frac{x}{y} \cdot \frac{1}{y}-\sin \lleft(1+\frac{y}{x}\right)\left(-\frac{y}{x^2}\right)]\\ +\left(-\frac{y}{x^2}\right) \left[\sin \frac{x}{y}+\cos \lleft(1+\frac{y}{x}\right)]\\ x \frac{\partial z}{\partial xx}=y \left[\frac{1}{y} \cos \frac{x}{y}+\frac{y}{x^2} \sin \lleft(1+\frac{y}{x}\right)]\\ -\frac{y}{x} \left[\sin \frac{x}{y}+\cos \lleft(1+\frac{y}{x}\right)]\\ \Rightarrow \quad x \frac{\partial z}{\partial x}=\cos \frac{x}{y}+\frac{y^2}{x^2} \sin \lleft(1+\frac{y}{x})-z \end{array} Similarly, y ∂z∂y=−cos⁡xy−y2x2sin⁡\l\frac{\partial z}{\partial y}=-\cos \frac{x}{y}-\frac{y^2}{x^2} \sin \lleft(1+\frac{y}{x})+zOnaddingEqs.(i)and(ii)xz On adding Eqs. (i) and (ii) x\frac{\partial z}{\partial x}=-$y \frac{\partial z}{\partial y}
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