use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$. $y^{2}=e^{x^{2}} +…

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $y^{2}$ using the chain - rule and $e^{x^{2}}+18x$ using the chain - rule and power rule. $2y\frac{dy}{dx}=2xe^{x^{2}} + 18$

Step2: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{2xe^{x^{2}}+18}{2y}=\frac{xe^{x^{2}} + 9}{y}$

Step3: Differentiate $\frac{dy}{dx}$ to find $\frac{d^{2}y}{dx^{2}}$

Use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u=xe^{x^{2}}+9$ and $v = y$. First, find $u'$: $u'=e^{x^{2}}+2x^{2}e^{x^{2}}$ Then, $\frac{d^{2}y}{dx^{2}}=\frac{(e^{x^{2}} + 2x^{2}e^{x^{2}})y-(xe^{x^{2}} + 9)\frac{dy}{dx}}{y^{2}}$ Substitute $\frac{dy}{dx}=\frac{xe^{x^{2}}+9}{y}$ into the above formula: $\frac{d^{2}y}{dx^{2}}=\frac{(e^{x^{2}}+2x^{2}e^{x^{2}})y-(xe^{x^{2}} + 9)\frac{xe^{x^{2}}+9}{y}}{y^{2}}=\frac{y^{2}(e^{x^{2}}+2x^{2}e^{x^{2}})-(xe^{x^{2}} + 9)^{2}}{y^{3}}$

Answer:

$\frac{dy}{dx}=\frac{xe^{x^{2}}+9}{y}$; $\frac{d^{2}y}{dx^{2}}=\frac{y^{2}(e^{x^{2}}+2x^{2}e^{x^{2}})-(xe^{x^{2}} + 9)^{2}}{y^{3}}$