જો z = frac{y}{x} left[ sin frac{x}{y} + cos | vedclass

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(B) આપેલ છે કે $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$.પ્રથમ, $\frac{\partial z}{\partial x}$ ની ગણતરી

Vedclass · Accepted Answer

$-y \frac{\partial z}{\partial y}$

Vedclass · Answer

(B) આપેલ છે કે $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$.પ્રથમ, $\frac{\partial z}{\partial x}$ ની ગણતરી કરીએ:$\frac{\partial z}{\partial x} = \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \frac{1}{y} - \sin \left( 1 + \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right) \right] - \frac{y}{x^2} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z \quad \dots (i)$ત્યારબાદ, $\frac{\partial z}{\partial y}$ ની ગણતરી કરીએ:$\frac{\partial z}{\partial y} = \frac{1}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] + \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$$y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right)$$y \frac{\partial z}{\partial y} = z - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right) \quad \dots (ii)$સમીકરણ $(i)$ અને $(ii)$ ની સરખામણી કરતા, આપણે જોઈ શકીએ છીએ કે $x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

જો $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ હોય, તો $x \frac{\partial z}{\partial x}$ કોના બરાબર થાય?

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