Question

use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^{2}y}{dx^{2}}$.
$y^{2}=5e^{x^{2}} + 12x$
$\frac{dy}{dx}=\frac{5xe^{x^{2}}+6}{y}$
$\frac{d^{2}y}{dx^{2}}=square$

Explanation:

Step1: Differentiate both sides of $\frac{dy}{dx}=\frac{5xe^{x^{2}} + 6}{y}$ with respect to $x$ using the quotient - rule.

The quotient - rule states that if $u$ and $v$ are functions of $x$ and $z=\frac{u}{v}$, then $z^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 5xe^{x^{2}}+6$, $u^\prime=5e^{x^{2}}+5x\cdot2x e^{x^{2}}=5e^{x^{2}}(1 + 2x^{2})$, and $v = y$, $v^\prime=\frac{dy}{dx}$. So, $\frac{d^{2}y}{dx^{2}}=\frac{(5e^{x^{2}}(1 + 2x^{2}))y-(5xe^{x^{2}} + 6)\frac{dy}{dx}}{y^{2}}$.

Step2: Substitute $\frac{dy}{dx}=\frac{5xe^{x^{2}} + 6}{y}$ into the above formula.

$\frac{d^{2}y}{dx^{2}}=\frac{5ye^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)\frac{5xe^{x^{2}}+6}{y}}{y^{2}}=\frac{5y^{2}e^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)^{2}}{y^{3}}$. Since $y^{2}=5e^{x^{2}}+12x$, we can also write it as $\frac{5(5e^{x^{2}}+12x)e^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)^{2}}{y^{3}}$.

Answer:

$\frac{5y^{2}e^{x^{2}}(1 + 2x^{2})-(5xe^{x^{2}} + 6)^{2}}{y^{3}}$