Question

If ${\log }_e(x+y)=4xy$. Find $\frac{d^{2}y}{d{x}^2}$ at $x=0$

Answer 1

Step 1: Solve the value of $y$ when $x=0$

${\log }_e(x+y)=4xy$

$\Rightarrow$ $x+y=e^{4xy}$ $...\left(i\right)$ $\left[\because {\log }_{e}a=b\Rightarrow e^b=a\right]$

Substitute value of $x=0$

$\Rightarrow$ $y=e^0$

$\Rightarrow$ $y=1$

Step 2: Solve for the first order derivative of given equation

Differentiating with respect to $x$ we get

$\Rightarrow$ $1+\frac{dy}{dx}=e^{4xy}\left(4x\frac{dy}{dx}+4y\right)$ $\left[\because \frac{d}{dx}{e}^{f\left(x,y\right)}=e^{f\left(x,y\right)}\times \frac{d}{dx}\left(f\left(x,y\right)\right)\right]$

$\Rightarrow$ $\frac{dy}{dx}\left(1-4x{e}^{4xy}\right)=4y{e}^{4xy}-1$

$\Rightarrow$ $\frac{dy}{dx}=\frac{4y{e}^{4xy}-1}{1-4x{e}^{4xy}}$

Substitute the values of $x=0,y=1$

$\Rightarrow$ $\frac{dy}{dx}=\frac{4\left(1\right)-1}{1-0}$

$\Rightarrow$ $\frac{dy}{dx}=3$

Step 3 : Solve for the second order derivative

$\frac{dy}{dx}=\frac{4y{e}^{4xy}-1}{1-4x{e}^{4xy}}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{4y\left(x+y\right)-1}{1-4x\left(x+y\right)}$ [From $\left(i\right)$]

$\Rightarrow$ $\frac{dy}{dx}=\frac{4y^2+4xy-1}{1-4x^2-4xy}$ $...\left(ii\right)$

Differentiating $\left(ii\right)$ with respect to $x$ we get

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{\left(1-4x^2-4xy\right)\left(8y{\displaystyle \frac{dy}{dx}}+4y+4x{\displaystyle \frac{dy}{dx}}\right)-\left(4y^2+4xy-1\right)\left(-8x-4y-4x{\displaystyle \frac{dy}{dx}}\right)}{{\left(1-4x^2-4xy\right)}^2}$ $\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{\left(1-4x^2-4xy\right)\left((4x+8y{\displaystyle )\frac{dy}{dx}}+4y\right)+\left(4y^2+4xy-1\right)\left(8x+4y+4x{\displaystyle \frac{dy}{dx}}\right)}{{\left(1-4x^2-4xy\right)}^2}$

Substitute the value of $x=0,y=1,\frac{dy}{dx}=3$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{\left(1-0-0\right)\left((0+8)\left(3\right)+4\right)+\left(4+0-1\right)\left(0+4+0\right)}{{\left(1-0-0\right)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{\left(28\right)+\left(3\right)\left(4\right)}{{\left(1\right)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=40$

Hence, the value of $\frac{d^{2}y}{d{x}^2}$ at $x=0$ is $40$.