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

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(B) दिया गया है $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$। सबसे पहले,$\frac{\partial z}{\partial x}$ की

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

Vedclass · Answer

(B) दिया गया है $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$। सबसे पहले,$\frac{\partial z}{\partial x}$ की गणना करें: $\frac{\partial z}{\partial x} = \frac{y}{x} \left[ \cos \frac{x}{y} \cdot \frac{1}{y} - \sin \left( 1 + \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right) \right] - \frac{y}{x^2} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ $x$ से गुणा करने पर: $x \frac{\partial z}{\partial x} = \cos \frac{x}{y} + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ $x \frac{\partial z}{\partial x} = \cos \frac{x}{y} + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z$ ... $(i)$ अब,$\frac{\partial z}{\partial y}$ की गणना करें: $\frac{\partial z}{\partial y} = \frac{1}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] + \frac{y}{x} \left[ \cos \frac{x}{y} \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$ $y$ से गुणा करने पर: $y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \frac{x}{y} - \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right)$ $y \frac{\partial z}{\partial y} = z - \cos \frac{x}{y} - \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right)$ ... $(ii)$ $(i)$ और $(ii)$ को जोड़ने पर: $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0$ अतः,$x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$।

यदि $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ है,तो $x \frac{\partial z}{\partial x}$ किसके बराबर है?

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निम्नलिखित अभिक्रिया में $B$ को पहचानें:
$H_4SiO_4$ $\xrightarrow[1000^{\circ}C]{-2H_2O} A$ $\xrightarrow[\Delta]{\text{Carbon}} B + CO$

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