implicit differentiation: problem 3 (1 point) find an equation of the line tangent to the curve defined by…

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:

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