Differentiation of trigonometry equations

greghunter · #1 $Old$ June 27th, 2009, 11:08 AM

Given that $y=\cos2x + sin2x$ , find $\frac{d^2y}{dx^2}$ and show that $\frac{d^2y}{dx^2} + 4y = 0$

Please help. I have tried to use the two trig calculus rules that I know but I can't seem to get them to work.

Plato · #2 $Old$ June 27th, 2009, 11:31 AM

We know that $\frac{{d^2 y}}{{dx^2 }} = - 4\cos (2x) - 4\sin (2x)$ .

Soroban · #3 $Old$ June 27th, 2009, 11:36 AM

Hello, Greg!

You must be differentiating incorrectly . . .

Quote:

Given that: . $y\:=\:\cos2x + \sin2x$ , .find $\frac{d^2y}{dx^2}$

and show that: . $\frac{d^2y}{dx^2} + 4y \:=\: 0$

We have: . $\begin{array}{ccc}y &=& \cos2x + \sin2x \\ \\ [-3mm] \dfrac{dy}{dx} &=& \text{-}2\sin2x + 2\cos2x \\ \\[-3mm] \dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin2x \end{array}$

Therefore: . $\begin{array}{ccccccc} \dfrac{d^2y}{dx^2} &=& \text{-}4\cos2x - 4\sin 2x \\ \\[-3mm] +4y\;\; &=& 4\cos2x + 4\sin2x \\ \hline \\[-4mm] \dfrac{d^2y}{dx^2} + 4y &=& 0\qquad\qquad\qquad \end{array}$

greghunter · #4 $Old$ June 27th, 2009, 12:28 PM

Can you tell me how you got from cos2x to -2sin2x?
I know that cosx = -sinx but the 2x confused me

Amer · #5 $Old$ June 27th, 2009, 01:16 PM

$\frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x))$ chain rule

$\frac{d}{dx}(cos(x^2))=2x^2(-sin(x^2))$

$\frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))$

greghunter · #6 $Old$ June 27th, 2009, 04:49 PM

Quote:

Originally Posted by Amer $View Post$

$\frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x))$ chain rule

$\frac{d}{dx}(cos(x^2))=2x^2(-sin(x^2))$

$\frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))$

thanks, i don't think i've been taught the chain rule yet