Question

implicit differentiation: problem 3 (1 point) find an equation of the line tangent to the curve defined by $x^{5}+4xy + y^{2}=17$ at the point $(1,2). y =$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}+4xy + y^{2}=17$ with respect to $x$.
Using the sum - rule, we have $\frac{d}{dx}(x^{3})+\frac{d}{dx}(4xy)+\frac{d}{dx}(y^{2})=\frac{d}{dx}(17)$.
For $\frac{d}{dx}(x^{3}) = 3x^{2}$ by the power - rule.
For $\frac{d}{dx}(4xy)$, using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 4x$ and $v = y$, we get $4y + 4x\frac{dy}{dx}$.
For $\frac{d}{dx}(y^{2})$, using the chain - rule, we get $2y\frac{dy}{dx}$. And $\frac{d}{dx}(17)=0$.
So, $3x^{2}+4y + 4x\frac{dy}{dx}+2y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ together:
$4x\frac{dy}{dx}+2y\frac{dy}{dx}=-3x^{2}-4y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4x + 2y)=-3x^{2}-4y$.
Then $\frac{dy}{dx}=\frac{-3x^{2}-4y}{4x + 2y}$.

Step3: Find the slope at the point $(1,2)$

Substitute $x = 1$ and $y = 2$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,2)}=\frac{-3(1)^{2}-4(2)}{4(1)+2(2)}=\frac{-3 - 8}{4 + 4}=\frac{-11}{8}$.

Step4: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,2)$ and $m =-\frac{11}{8}$.
$y - 2=-\frac{11}{8}(x - 1)$.
Expand and solve for $y$:
$y-2=-\frac{11}{8}x+\frac{11}{8}$.
$y=-\frac{11}{8}x+\frac{11}{8}+2$.
$y=-\frac{11}{8}x+\frac{11 + 16}{8}$.
$y=-\frac{11}{8}x+\frac{27}{8}$.

Answer:

$y =-\frac{11}{8}x+\frac{27}{8}$