Question

consider the following. 9x^2 - y^5 = 8 (a) find y by implicit differentiation. y = (b) solve the equation explicitly for y and differentiate to get y in terms of x. y = (c) check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a). y =

Explanation:

Step1: Differentiate both sides for (a)

Differentiate $9x^{2}-y^{5}=8$ with respect to $x$. The derivative of $9x^{2}$ with respect to $x$ is $18x$ using the power - rule. For $-y^{5}$, by the chain - rule, it is $-5y^{4}y'$. The derivative of the constant 8 is 0. So we have $18x-5y^{4}y' = 0$.

Step2: Solve for $y'$ in (a)

Rearrange $18x-5y^{4}y' = 0$ to get $y'=\frac{18x}{5y^{4}}$.

Step3: Solve for $y$ in (b)

First, solve $9x^{2}-y^{5}=8$ for $y$. We get $y=(9x^{2}-8)^{\frac{1}{5}}$.

Step4: Differentiate $y$ in (b)

Using the chain - rule, if $y=(9x^{2}-8)^{\frac{1}{5}}$, then $y'=\frac{1}{5}(9x^{2}-8)^{-\frac{4}{5}}\cdot18x=\frac{18x}{5(9x^{2}-8)^{\frac{4}{5}}}$. Since $y=(9x^{2}-8)^{\frac{1}{5}}$, we can rewrite it as $y'=\frac{18x}{5y^{4}}$.

Step5: Consistency check for (c)

Substitute $y=(9x^{2}-8)^{\frac{1}{5}}$ into the solution of part (a) $y'=\frac{18x}{5y^{4}}$. We get $y'=\frac{18x}{5(9x^{2}-8)^{\frac{4}{5}}}$, which is the same as the result of part (b).

Answer:

(a) $\frac{18x}{5y^{4}}$
(b) $\frac{18x}{5y^{4}}$
(c) $\frac{18x}{5y^{4}}$