given the equation 7y² - x³ = 5, find d²y/dx² by implicit differentiation.

Explanation:

Step1: Differentiate both sides w.r.t. $x$

Differentiating $7y^{2}-x^{3}=5$ with respect to $x$ using the chain - rule for the $y$ terms. The derivative of $7y^{2}$ with respect to $x$ is $14y\frac{dy}{dx}$ (by chain - rule: $\frac{d}{dx}(u^{2}) = 2u\frac{du}{dx}$, here $u = y$), and the derivative of $-x^{3}$ with respect to $x$ is $-3x^{2}$, and the derivative of the constant $5$ is $0$. So we have: $14y\frac{dy}{dx}-3x^{2}=0$.

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

Isolate $\frac{dy}{dx}$: $14y\frac{dy}{dx}=3x^{2}$, so $\frac{dy}{dx}=\frac{3x^{2}}{14y}$.

Step3: Differentiate $\frac{dy}{dx}$ with respect to $x$ to get $\frac{d^{2}y}{dx^{2}}$

Using the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = 3x^{2}$, $u'=6x$, $v = 14y$, and $v'=14\frac{dy}{dx}$. $\frac{d^{2}y}{dx^{2}}=\frac{6x\cdot14y-3x^{2}\cdot14\frac{dy}{dx}}{(14y)^{2}}$. Substitute $\frac{dy}{dx}=\frac{3x^{2}}{14y}$ into the above formula: [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{84xy-3x^{2}\cdot14\cdot\frac{3x^{2}}{14y}}{196y^{2}}\ &=\frac{84xy-\frac{42x^{4}}{y}}{196y^{2}}\ &=\frac{\frac{84xy^{2}-42x^{4}}{y}}{196y^{2}}\ &=\frac{84xy^{2}-42x^{4}}{196y^{3}}\ &=\frac{42(2xy^{2}-x^{4})}{196y^{3}}\ &=\frac{3(2xy^{2}-x^{4})}{14y^{3}} \end{align*} ]

Answer:

$\frac{3(2xy^{2}-x^{4})}{14y^{3}}$