Differentiation of Sinusoidal Functions

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Apr 13th 2010, 03:50 PM
Eixo

Differentiation of Sinusoidal Functions

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

Shown work would be appreciated, thank you.
Apr 13th 2010, 04:04 PM
skeeter

Quote:

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.

why don't you show the work of finding $\frac{dy}{dx}$ , first?
Apr 13th 2010, 04:06 PM
Eixo

Quote:

Originally Posted by skeeter

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

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 $\frac{dy}{dx}$
Apr 13th 2010, 04:10 PM
skeeter

Quote:

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 $\frac{dy}{dx}$

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

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

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

this is a very basic process. give it a go ...
Apr 13th 2010, 04:30 PM
Eixo

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

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

I then subbed in $-\pi$ (typo from original post, will edit).

$=((-\pi)^2(2(1)))+(2(-\pi)(0))$
$=2\pi^2$

$y$ @ $-\pi = (-\pi)^2 \cdot sin(2(-\pi))<br /> = 0$

Slope = Slope

$2\pi^2=\frac{y-0}{x+\pi}$ (Cross multiply)
$2\pi^2x+2\pi^3 = y$
Apr 13th 2010, 04:37 PM
skeeter

Quote:

Originally Posted by Eixo

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

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

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

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

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

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

...
Apr 13th 2010, 04:40 PM
Eixo

Sorry, will correct. My answer remains unchanged though.

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