If $x=e^{(y+e)^{(y+e)}(y+\ldots \ldots \infty)}$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

Here $x=e^{y+x}$ Differentiating w.r.t. $\mathrm{x}$ $\begin{array}{l} 1=e^{y+x}\left(\frac{d y}{d x}+1ight) \Rightarrow 1=e^{y+x} \frac{d y}{d x}+e^{y+x} \ 	herefore 1=x \cdot \frac{d y}{d x}+x \ \frac{d y}{d x}=\frac{1-x}{x} \end{array}$

If $x=e^{(y+e)^{(y+e)}(y+\ldots \ldots \infty)}$, then $\frac{d y}{d x}=$

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