Differentiation of Sinusoidal Functions

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.

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Originally Posted by Eixo

Find the equation of the line that is tangent to y = x^2sin(2x) at x = pi.


Shown work would be appreciated, thank you.

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why don't you show the work of finding $\displaystyle \frac{dy}{dx}$, first?

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Originally Posted by skeeter

why don't you show the work of finding $\displaystyle \frac{dy}{dx}$, first?

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That's unfortunately where I've run into my trouble. I know how to do the rest of the question, however I am unsure on how to find $\displaystyle \frac{dy}{dx}$

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Originally Posted by Eixo

That's unfortunately where I've run into my trouble. I know how to do the rest of the question, however I am unsure on how to find $\displaystyle \frac{dy}{dx}$

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product rule + chain rule for $\displaystyle \sin(2x)$ ...

$\displaystyle \frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$

where $\displaystyle u = x^2$ and $\displaystyle v = \sin(2x)$


this is a very basic process. give it a go ...

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)
$\displaystyle 2\pi^2x+2\pi^3 = y$

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Originally Posted by Eixo

Thank you for the help, it was the mental roadblock that I was struggling to get past.

$\displaystyle \frac{dy}{dx}$=(x^2)(2cos(2x))+(2x)(2sin(2x))

close ... $\displaystyle \textcolor{red}{\frac{dy}{dx} = x^2 \cdot 2\cos(2x) + \sin(2x) \cdot 2x}$

now sub in $\displaystyle \pi$ for x to find the slope.

use of the point-slope form is a bit easier ...

$\displaystyle \textcolor{red}{y - y_1 = m(x - x_1)}$

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Sorry, will correct. My answer remains unchanged though.