find an equation of the tangent line to the curve at the point (0, 1). y=(1 + 3x)^10 y= need help? read it…

Explanation:

Step1: Find the derivative using chain - rule

Let $u = 1+3x$, then $y = u^{10}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=10u^{9}$ and $\frac{du}{dx}=3$. So $\frac{dy}{dx}=10(1 + 3x)^{9}\cdot3=30(1 + 3x)^{9}$.

Step2: Evaluate the derivative at the given point

We want to find the slope of the tangent line at $x = 0$. Substitute $x = 0$ into $\frac{dy}{dx}$. When $x = 0$, $\frac{dy}{dx}\big|_{x = 0}=30(1+3\cdot0)^{9}=30$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,1)$ and $m = 30$. $y-1=30(x - 0)$.

Step4: Simplify the equation

$y-1 = 30x$, so $y=30x + 1$.

Answer:

$y = 30x+1$