Question

question let y be defined implicitly by the equation 6x^5 + 6y^4=-45xy. use implicit differentiation to evaluate dy/dx at the point (-1,2). (submit an exact answer.) provide your answer below:

Explanation:

Step1: Differentiate both sides

Differentiate $6x^{5}+6y^{4}=- 45xy$ with respect to $x$.
Using the sum - rule, we have $\frac{d}{dx}(6x^{5})+\frac{d}{dx}(6y^{4})=\frac{d}{dx}(-45xy)$.
For $\frac{d}{dx}(6x^{5}) = 30x^{4}$ by the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
For $\frac{d}{dx}(6y^{4})$, by the chain - rule, it is $24y^{3}\frac{dy}{dx}$.
For $\frac{d}{dx}(-45xy)$, by the product - rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ where $u=-45x$ and $v = y$, we get $-45y-45x\frac{dy}{dx}$.
So, $30x^{4}+24y^{3}\frac{dy}{dx}=-45y - 45x\frac{dy}{dx}$.

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

Move all terms with $\frac{dy}{dx}$ to one side:
$24y^{3}\frac{dy}{dx}+45x\frac{dy}{dx}=-45y - 30x^{4}$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(24y^{3}+45x)=-45y - 30x^{4}$.
Then $\frac{dy}{dx}=\frac{-45y - 30x^{4}}{24y^{3}+45x}$.

Step3: Substitute the point $(-1,2)$

Substitute $x=-1$ and $y = 2$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{x=-1,y = 2}=\frac{-45\times2-30\times(-1)^{4}}{24\times2^{3}+45\times(-1)}$.
First, calculate the numerator: $-45\times2-30\times(-1)^{4}=-90 - 30=-120$.
Then, calculate the denominator: $24\times2^{3}+45\times(-1)=24\times8-45=192 - 45 = 147$.
So, $\frac{dy}{dx}=\frac{-120}{147}=-\frac{40}{49}$.

Answer:

$-\frac{40}{49}$