# If f(x)=5x(sinx)(cosx), then what does f'(x) equal?

## No idea how to take the derivative of this. I thought it would be 5(cosx)(-sinx), but my online homework program tells me this is wrong. How do I find the derivative here?

Mar 6, 2018

f'(x)=5(cos(x)sin(x)+x(cos^2(x)-sin^2(x))

#### Explanation:

We want to find the derivative of

$f \left(x\right) = 5 x \sin \left(x\right) \cos \left(x\right)$

With repeated use of the product, if $f = g \cdot h \cdot q$ then

$f ' = \left(g \cdot h \cdot q\right) '$$= \left(g \cdot h\right) ' \cdot q + g \cdot h \cdot q '$

$= \left(g ' \cdot h + g \cdot h '\right) \cdot q + g \cdot h \cdot q '$

$= g ' \cdot h \cdot q + g \cdot h ' \cdot q + g \cdot h \cdot q '$

Here

• $g = 5 x \implies g ' = 5$

• $h = \cos \left(x\right) \implies h ' = - \sin \left(x\right)$

• $q = \sin \left(x\right) \implies q ' = \cos \left(x\right)$

$f ' \left(x\right) = 5 \cos \left(x\right) \sin \left(x\right) - 5 x \sin \left(x\right) \sin \left(x\right) + 5 x \cos \left(x\right) \cos \left(x\right)$

$= 5 \cos \left(x\right) \sin \left(x\right) - 5 x {\sin}^{2} \left(x\right) + 5 x {\cos}^{2} \left(x\right)$

=5(cos(x)sin(x)+x(cos^2(x)-sin^2(x))