3. ** find the equation of the tangent line at the given point: y³ + 4x = 3y + 4x² at (1,0)

Explanation:

Step1: Differentiate implicitly

Differentiate $y^{3}+4x = 3y + 4x^{2}$ with respect to $x$. Using the chain - rule for terms involving $y$: $\frac{d}{dx}(y^{3})=3y^{2}y'$, $\frac{d}{dx}(4x) = 4$, $\frac{d}{dx}(3y)=3y'$, $\frac{d}{dx}(4x^{2}) = 8x$. So, $3y^{2}y'+4 = 3y'+8x$.

Step2: Solve for $y'$

Rearrange the terms to isolate $y'$: $3y^{2}y'-3y'=8x - 4$. Factor out $y'$: $y'(3y^{2}-3)=8x - 4$. Then $y'=\frac{8x - 4}{3y^{2}-3}$.

Step3: Find the slope at the point $(1,0)$

Substitute $x = 1$ and $y = 0$ into $y'$: $y'=\frac{8\times1 - 4}{3\times0^{2}-3}=\frac{4}{-3}=-\frac{4}{3}$.

Step4: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,0)$ and $m =-\frac{4}{3}$. $y-0=-\frac{4}{3}(x - 1)$. Simplify to get $y=-\frac{4}{3}x+\frac{4}{3}$ or $4x+3y = 4$.

Answer:

$y=-\frac{4}{3}x+\frac{4}{3}$ (or $4x + 3y=4$)