Question

question let y be defined implicitly by the equation 8x^3 - 7y^3+5y = 79. use implicit differentiation to find $\frac{dy}{dx}$. provide your answer below: $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $8x^{3}-7y^{3}+5y = 79$ with respect to $x$. The derivative of $8x^{3}$ with respect to $x$ is $24x^{2}$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. For $-7y^{3}$, by the chain - rule, it is $-21y^{2}\frac{dy}{dx}$. The derivative of $5y$ with respect to $x$ is $5\frac{dy}{dx}$, and the derivative of the constant $79$ is $0$. So we have $24x^{2}-21y^{2}\frac{dy}{dx}+5\frac{dy}{dx}=0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $-21y^{2}\frac{dy}{dx}+5\frac{dy}{dx}=-24x^{2}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(-21y^{2}+5)=-24x^{2}$. Then solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{24x^{2}}{21y^{2}-5}$.

Answer:

$\frac{24x^{2}}{21y^{2}-5}$