# How do you find the second derivative of y=2sin3x-5sin6x?

Feb 23, 2015

First find the first derivative using the chain rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 \cos \left(3 x\right) - 30 \cos \left(6 x\right)$

Now take the derivative again to find the second derivative

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 18 \sin \left(3 x\right) + 180 \sin \left(6 x\right)$

Feb 23, 2015

Hello,

Answer $y ' ' = - 18 \sin \left(3 x\right) + 180 \sin \left(6 x\right)$.

• First, you calculate the derivative :
$y ' = 2 \cos \left(3 x\right) \setminus \times 3 - 5 \cos \left(6 x\right) \setminus \times 6$.

I used $\frac{d}{\mathrm{dx}} \sin \left(n x\right) = \cos \left(n x\right) \setminus \times n$.

• You can simplify :
$y ' = 6 \cos \left(3 x\right) - 30 \cos \left(6 x\right)$.

• Now, you calculate the second derivative :
$y ' ' = - 6 \sin \left(3 x\right) \setminus \times 3 + 30 \sin \left(6 x\right) \setminus \times 6$

I used $\frac{d}{\mathrm{dx}} \cos \left(n x\right) = - \sin \left(n x\right) \setminus \times n$.

You can also simplify to obtain the result.