What is the derivative of # f(x)=cos^3x#?

Question

Isabella Ackerman · Accepted Answer

#f'(x)=-3sinxcos^2x# #"differentiate using the "color(blue)"chain rule"# #"Given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"# #f(x)=cos^3x=(cosx)^3# #rArrf'(x)=3(cosx)^2xxd/dx(cosx)# #color(white)(rArrf'(x))=3cos^2x xx-sinx# #color(white)(rArrf'(x))=-3sinxcos^2x#

Emily Ammerman · Answer

undefined To find the derivative of $ f(x) = \cos^3(x) $, you can apply the chain rule of differentiation.

The chain rule states that if $ u(x) $ is differentiable at $ x $ and $ f(x) $ is differentiable at $ u $, then the composite function $ f(u(x)) $ is differentiable at $ x $, and its derivative is given by $ f'(u(x)) \cdot u'(x) $.

First, differentiate the outer function $ \cos^3(x) $ with respect to its inner function $ \cos(x) $. The derivative of $ \cos^3(x) $ with respect to $ \cos(x) $ is $ 3\cos^2(x) $.

Then, differentiate the inner function $ \cos(x) $ with respect to $ x $. The derivative of $ \cos(x) $ is $ -\sin(x) $.

Finally, apply the chain rule by multiplying the derivative of the outer function by the derivative of the inner function:

$$ f'(x) = 3\cos^2(x) \cdot (-\sin(x)) $$

So, the derivative of $ f(x) = \cos^3(x) $ is $ -3\cos^2(x)\sin(x) $.