# Thread: Differentiation of trigonometry equations

1. ## Differentiation of trigonometry equations

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

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.

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

3. Hello, Greg!

You must be differentiating incorrectly . . .

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

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

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

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

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

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

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

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

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

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

$\displaystyle \frac{d}{dx}(cos(5x-1))=\frac{d}{dx}(5x-1) (-sin(5x-1))=5 (-sin(5x-1))$
thanks, i don't think i've been taught the chain rule yet