Question

implicit differentiation: problem 3 (1 point) results for this submission the answer is not correct. find an equation of the line tangent to the curve defined by $x^{5}+4xy + y^{3}=17$ at the point $(1,2).$ $y =-\frac{11}{8}x+\frac{27}{8}$ preview my answers submit answers your score was recorded. your score was successfully sent to canvas. you have attempted this problem 1 time. you received a score of 0% for this attempt. your overall recorded score is 0%. you have unlimited attempts remaining.

Explanation:

Step1: Differentiate both sides

Differentiate $x^{5}+4xy + y^{3}=17$ with respect to $x$.
Using sum - rule and product - rule, we have:
$\frac{d}{dx}(x^{5})+\frac{d}{dx}(4xy)+\frac{d}{dx}(y^{3})=\frac{d}{dx}(17)$.
For $\frac{d}{dx}(x^{5}) = 5x^{4}$, for $\frac{d}{dx}(4xy)$ using product - rule $(uv)^\prime=u^\prime v + uv^\prime$ where $u = 4x$ and $v = y$, we get $4y+4x\frac{dy}{dx}$, and for $\frac{d}{dx}(y^{3})$ using chain - rule we get $3y^{2}\frac{dy}{dx}$, and $\frac{d}{dx}(17)=0$.
So, $5x^{4}+4y + 4x\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ on one side:
$4x\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=-5x^{4}-4y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4x + 3y^{2})=-5x^{4}-4y$.
Then $\frac{dy}{dx}=\frac{-5x^{4}-4y}{4x + 3y^{2}}$.

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

Substitute $x = 1$ and $y = 2$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,2)}=\frac{-5(1)^{4}-4(2)}{4(1)+3(2)^{2}}=\frac{-5 - 8}{4 + 12}=\frac{-13}{16}$.

Step4: Use point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,2)$ and $m=\frac{-13}{16}$.
$y - 2=\frac{-13}{16}(x - 1)$.
Expand: $y-2=-\frac{13}{16}x+\frac{13}{16}$.
$y=-\frac{13}{16}x+\frac{13}{16}+2=-\frac{13}{16}x+\frac{13 + 32}{16}=-\frac{13}{16}x+\frac{45}{16}$.

Answer:

$y =-\frac{13}{16}x+\frac{45}{16}$