# Differentiation of trigonometry equations

• Jun 27th 2009, 09:08 AM
greghunter
Differentiation of trigonometry equations
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.
• Jun 27th 2009, 09:31 AM
Plato
We know that $\frac{{d^2 y}}{{dx^2 }} = - 4\cos (2x) - 4\sin (2x)$.
• Jun 27th 2009, 09:36 AM
Soroban
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]

\end{array}$

• Jun 27th 2009, 10:28 AM
greghunter
Can you tell me how you got from cos2x to -2sin2x?
I know that cosx = -sinx but the 2x confused me
• Jun 27th 2009, 11:16 AM
Amer
$\frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x))$ chain rule

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

$\frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))$
• Jun 27th 2009, 02:49 PM
greghunter
Quote:

Originally Posted by Amer
$\frac{d}{dx}(cos(f(x)))=f'(x)(-sin(f(x))$ 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