Find the equation of the line that is tangent to $\displaystyle y = x^2sin(2x)$ at $\displaystyle x = -\pi.$
Shown work would be appreciated, thank you.
Thank you for the help, it was the mental roadblock that I was struggling to get past.
$\displaystyle \frac{dy}{dx}=x^2 \cdot 2cos(2x)+2x \cdot 2sin(2x)$
I then subbed in $\displaystyle -\pi$ (typo from original post, will edit).
$\displaystyle =((-\pi)^2(2(1)))+(2(-\pi)(0))$
$\displaystyle =2\pi^2$
$\displaystyle y $@$\displaystyle -\pi = (-\pi)^2 \cdot sin(2(-\pi))
= 0$
Slope = Slope
$\displaystyle 2\pi^2=\frac{y-0}{x+\pi}$ (Cross multiply)
$\displaystyle 2\pi^2x+2\pi^3 = y$