Question

find y and the slope of the tangent line to the graph of (3x - 2y)^5 = 2y^2 - 9 at the point (1,2).
y =
$y|_{(1,2)}=$ (simplify your answer.)

Explanation:

Step1: Differentiate both sides with chain - rule

Differentiate $(3x - 2y)^5$ using the chain - rule. Let $u = 3x-2y$, then $\frac{d}{dx}(u^5)=5u^4\frac{du}{dx}=5(3x - 2y)^4(3 - 2y')$. Differentiate $2y^2-9$ with respect to $x$: $\frac{d}{dx}(2y^2-9)=4yy'$. So, $5(3x - 2y)^4(3 - 2y')=4yy'$.

Step2: Expand the left - hand side

Expand $5(3x - 2y)^4(3 - 2y')$ to get $15(3x - 2y)^4-10(3x - 2y)^4y'=4yy'$.

Step3: Isolate $y'$ terms

Move all terms with $y'$ to one side: $15(3x - 2y)^4=4yy'+10(3x - 2y)^4y'$. Factor out $y'$: $15(3x - 2y)^4=y'(4y + 10(3x - 2y)^4)$.

Step4: Solve for $y'$

$y'=\frac{15(3x - 2y)^4}{4y + 10(3x - 2y)^4}$.

Step5: Find the slope at the point $(1,2)$

Substitute $x = 1$ and $y = 2$ into $y'$:
\[

$$\begin{align*} y'\big|_{(1,2)}&=\frac{15(3\times1 - 2\times2)^4}{4\times2+10(3\times1 - 2\times2)^4}\\ &=\frac{15(-1)^4}{8 + 10(-1)^4}\\ &=\frac{15}{8 + 10}\\ &=\frac{15}{18}\\ &=\frac{5}{6} \end{align*}$$

\]

Answer:

$y'=\frac{15(3x - 2y)^4}{4y + 10(3x - 2y)^4}$; $y'\big|_{(1,2)}=\frac{5}{6}$