If {x^x}{y^y}{z^z} = c,then {{partial z} over | vedclass

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Question

(B) Given the equation: ${x^x}{y^y}{z^z} = c$ Taking the natural logarithm on both sides: $\log({x^x}{y^y}{z^z}) = \log c$ Using logarithmic propertie

Vedclass · Accepted Answer

$ - {{1 + \log x} \over {1 + \log z}}$

Vedclass · Answer

(B) Given the equation: ${x^x}{y^y}{z^z} = c$ Taking the natural logarithm on both sides: $\log({x^x}{y^y}{z^z}) = \log c$ Using logarithmic properties: $x\log x + y\log y + z\log z = \log c$ .....$(i)$ Here,$x$ and $y$ are independent variables,and $z$ is a function of $x$ and $y$. Differentiating equation $(i)$ partially with respect to $x$: $\frac{\partial}{\partial x}(x\log x) + \frac{\partial}{\partial x}(y\log y) + \frac{\partial}{\partial x}(z\log z) = \frac{\partial}{\partial x}(\log c)$ $(1 \cdot \log x + x \cdot \frac{1}{x}) + 0 + (1 \cdot \log z + z \cdot \frac{1}{z}) \frac{\partial z}{\partial x} = 0$ $(1 + \log x) + (1 + \log z) \frac{\partial z}{\partial x} = 0$ Rearranging the terms to solve for $\frac{\partial z}{\partial x}$: $(1 + \log z) \frac{\partial z}{\partial x} = -(1 + \log x)$ $\frac{\partial z}{\partial x} = -\frac{1 + \log x}{1 + \log z}$ Thus,the correct option is $B$.

If ${x^x}{y^y}{z^z} = c$,then ${{\partial z} \over {\partial x}} = $

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