If sinhx+y=logx+y then dydx2=

Solution:

The given equation is $\sinh (x + y) = \log (x + y)$

Differentiate both sides

$\begin{array}{rcl} \cos h (x + y) [1 + \frac{d y}{d x}] & = & \frac{1}{x + y} [1 + \frac{d y}{d x}] \\ [1 + \frac{d y}{d x}] (\cos h (x + y) - \frac{1}{x + y}) & = & 0 \end{array}$

It implies that $1 + \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - 1$

Therefore, ${(\frac{d y}{d x})}^{2} = 1$

If $\sinh (x + y) = \log (x + y)$ then ${(\frac{dy}{dx})}^{2} =$

Solution:

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