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|_(1,2) = (simplify your answer.)

Explanation:

Step1: Differentiate both sides implicitly

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

Step2: Expand the left - hand side

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

Step3: Isolate $y'$ terms

$15(3x - 2y)^4=4yy'+10(3x - 2y)^4y'$.
Factor out $y'$ on the right - hand side: $15(3x - 2y)^4=y'(4y + 10(3x - 2y)^4)$.

Step4: Solve for $y'$

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

Step5: Substitute $x = 1$ and $y = 2$

Substitute $x = 1$ and $y = 2$ into $y'$:
First, calculate $(3x - 2y)$ when $x = 1$ and $y = 2$: $3\times1-2\times2=-1$.
$y'=\frac{15(-1)^4}{4\times2+10(-1)^4}=\frac{15}{8 + 10}=\frac{15}{18}=\frac{5}{6}$.

Answer:

$\frac{5}{6}$