If y=y(x) is an implicit function of x such t | vedclass

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(D) Given $\log _{e}(x+y)=4 x y$. At $x=0$,$\log _{e}(y)=0$,which implies $y=1$.Differentiating both sides with respect to $x$:$\frac{1}{x+y} \left(1+

Vedclass · Accepted Answer

$40$

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(D) Given $\log _{e}(x+y)=4 x y$. At $x=0$,$\log _{e}(y)=0$,which implies $y=1$.Differentiating both sides with respect to $x$:$\frac{1}{x+y} \left(1+\frac{d y}{d x}\right) = 4y + 4x \frac{d y}{d x}$.At $x=0$ and $y=1$:$\frac{1}{1} \left(1+\frac{d y}{d x}\right) = 4(1) + 4(0) \frac{d y}{d x} \Rightarrow 1+\frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = 3$.Now,differentiate $1+\frac{d y}{d x} = (x+y)(4y + 4x \frac{d y}{d x})$:$\frac{d^{2} y}{d x^{2}} = (1+\frac{d y}{d x})(4y + 4x \frac{d y}{d x}) + (x+y)(4 \frac{d y}{d x} + 4 \frac{d y}{d x} + 4x \frac{d^{2} y}{d x^{2}})$.Substitute $x=0, y=1, \frac{d y}{d x}=3$:$\frac{d^{2} y}{d x^{2}} = (1+3)(4(1) + 0) + (0+1)(4(3) + 4(3) + 0)$.$\frac{d^{2} y}{d x^{2}} = (4)(4) + (1)(24) = 16 + 24 = 40$.

If $y=y(x)$ is an implicit function of $x$ such that $\log _{e}(x+y)=4 x y$,then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is equal to .... .

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