Question

question 12 · 1 point
let y be defined implicitly by the equation
(8x + 3y)^3 = 9x^3 - 3y^2.
use implicit differentiation to find $\frac{dy}{dx}$.
provide your answer below:
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate left - hand side

Differentiate $(8x + 3y)^3$ using chain - rule. Let $u=8x + 3y$, then $\frac{d}{dx}(u^3)=3u^2\frac{du}{dx}$. Here, $\frac{du}{dx}=8 + 3\frac{dy}{dx}$, so $\frac{d}{dx}(8x + 3y)^3=3(8x + 3y)^2(8 + 3\frac{dy}{dx})$.

Step2: Differentiate right - hand side

Differentiate $9x^3-3y^2$ with respect to $x$. $\frac{d}{dx}(9x^3)=27x^2$ and $\frac{d}{dx}(-3y^2)=-6y\frac{dy}{dx}$ using chain - rule.

Step3: Set derivatives equal

Set the derivatives of the left - hand side and right - hand side equal:
\[3(8x + 3y)^2(8 + 3\frac{dy}{dx})=27x^2-6y\frac{dy}{dx}\]

Step4: Expand left - hand side

Expand $3(8x + 3y)^2(8 + 3\frac{dy}{dx})$:
\[3(8x + 3y)^2\times8+3(8x + 3y)^2\times3\frac{dy}{dx}=27x^2-6y\frac{dy}{dx}\]
\[24(8x + 3y)^2 + 9(8x + 3y)^2\frac{dy}{dx}=27x^2-6y\frac{dy}{dx}\]

Step5: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side:
\[9(8x + 3y)^2\frac{dy}{dx}+6y\frac{dy}{dx}=27x^2 - 24(8x + 3y)^2\]

Step6: Factor out $\frac{dy}{dx}$

\[\frac{dy}{dx}(9(8x + 3y)^2+6y)=27x^2 - 24(8x + 3y)^2\]

Step7: Solve for $\frac{dy}{dx}$

\[\frac{dy}{dx}=\frac{27x^2 - 24(8x + 3y)^2}{9(8x + 3y)^2+6y}\]

Answer:

\(\frac{27x^2 - 24(8x + 3y)^2}{9(8x + 3y)^2+6y}\)