How do you differentiate g(x) = sin(3x)sin(6x) using the product rule?

Feb 27, 2017

$g ' \left(x\right) = 3 \sin 6 x \cos 3 x + 6 \sin 3 x \cos 6 x$

Explanation:

The product rule says that if

$f \left(x\right) = a \left(x\right) \cdot b \left(x\right)$

then

$f ' \left(x\right) = a ' \left(x\right) b \left(x\right) + b ' \left(x\right) a \left(x\right)$

In the above example,

$g \left(x\right) = \sin \left(3 x\right) \sin \left(6 x\right)$,

so

$a \left(x\right) = \sin \left(3 x\right) , b \left(x\right) = \sin \left(6 x\right)$

Now we have our two separate functions, we can differentiate them. However, these functions have insides ($3 x$ or $6 x$) and outsides ($\sin$), so to differentiate them we multiply the derivative of the inside by the derivative of the outside with the inside left the same. In other words,

$\frac{d}{\mathrm{dx}} \sin \left(a x + b\right) = a \cos \left(a x + b\right)$

so

$a \left(x\right) = \sin \left(3 x\right) , b \left(x\right) = \sin \left(6 x\right)$

$a ' \left(x\right) = 3 \cos \left(3 x\right) , b ' \left(x\right) = 6 \cos \left(6 x\right)$

Therefore the derivative of the whole thing, if you remember back to the product rule equation, is

$g ' \left(x\right) = 3 \cos \left(3 x\right) \cdot \sin \left(6 x\right) + 6 \cos \left(6 x\right) \cdot \sin \left(3 x\right)$

$g ' \left(x\right) = 3 \sin 6 x \cos 3 x + 6 \sin 3 x \cos 6 x$