# How do you find the second derivative of y=Acos(Bx)?

Aug 19, 2017

Treat it like the product of two functions: $f \left(x\right) \cdot g \left(x\right)$
where $f \left(x\right) = A$ (just a constant) and $g \left(x\right) = \cos \left(B x\right)$

so $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 0$.
and $\left(\mathrm{dg} \frac{x}{\mathrm{dx}}\right) = - B \sin \left(B X\right)$ (used the chain rule here.)

So, by the product rule, the deriv. of the product $f \left(x\right) \cdot g \left(x\right) =$

$\left(\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right) g \left(x\right) + f \left(x\right) \left(\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}\right)$

in this case, that works out to $- A B \sin \left(B X\right)$

Now, just do it again:

$\frac{d}{\mathrm{dx}} \left(- A B \sin \left(B X\right)\right) = - A {B}^{2} \cos \left(B X\right)$ (by the chain rule)

GOOD LUCK!