use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$. $y^{2}=5e^{x^{2}} +…

Explanation:

Step1: Differentiate both sides with respect to x

Differentiate $y^{2}$ using the chain - rule ($\frac{d}{dx}(y^{2}) = 2y\frac{dy}{dx}$), differentiate $5e^{x^{2}}$ using the chain - rule ($\frac{d}{dx}(5e^{x^{2}})=5e^{x^{2}}\cdot2x = 10xe^{x^{2}}$), and $\frac{d}{dx}(12x)=12$. So, $2y\frac{dy}{dx}=10xe^{x^{2}} + 12$.

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

Divide both sides of the equation $2y\frac{dy}{dx}=10xe^{x^{2}} + 12$ by $2y$ to get $\frac{dy}{dx}=\frac{10xe^{x^{2}}+12}{2y}=\frac{5xe^{x^{2}} + 6}{y}$.

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 $u = 5xe^{x^{2}}+6$ and $v = y$. First, find $u'$: $\frac{d}{dx}(5xe^{x^{2}}+6)=5e^{x^{2}}+5x\cdot2xe^{x^{2}}=5e^{x^{2}}(1 + 2x^{2})$. And $v'=\frac{dy}{dx}=\frac{5xe^{x^{2}}+6}{y}$. Then $\frac{d^{2}y}{dx^{2}}=\frac{(5e^{x^{2}}(1 + 2x^{2}))y-(5xe^{x^{2}} + 6)\frac{dy}{dx}}{y^{2}}$. Substitute $\frac{dy}{dx}=\frac{5xe^{x^{2}}+6}{y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{y\cdot5e^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)\frac{5xe^{x^{2}}+6}{y}}{y^{2}}\ &=\frac{y^{2}\cdot5e^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)^{2}}{y^{3}} \end{align*} ] Since $y^{2}=5e^{x^{2}}+12x$, we can also express it in terms of $x$ only after substitution. But for the first - order derivative:

Answer:

$\frac{dy}{dx}=\frac{5xe^{x^{2}}+6}{y}$