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How do you differentiate #3sin^2(3x) # using the chain rule?

Answer 1

Charles Anctil

#18sin(3x)cos(3x)#

chain rule for 3 functions is #d/dx(f(g(h(x))))=f'(g(h(x)))*d/dx(g(h(x)))=f'(g(h(x)))g'(h(x))h'(x)#

in this problem, #f(x)=3x^2#, #g(x)=sin(x)#, and #h(x)=3x# that means: #f'(x)=6x#, #g'(x)=cos(x)#, and #h(x)=3#

so #d/dx(3sin^2(3x))=6(sin(3x))*cos(3x)*3# #=18sin(3x)cos(3x)#

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Answer 2

Isaiah Allen

To differentiate ( 3\sin^2(3x) ) using the chain rule, you apply the chain rule which states that ( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) ).

Let ( u = 3x ) and ( v = \sin(u) ).

Now, ( \frac{du}{dx} = 3 ) and ( \frac{dv}{du} = 2\sin(u)\cos(u) ) (using the chain rule for ( \sin^2(u) )).

Therefore, ( \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} = 2\sin(3x)\cos(3x) \cdot 3 = 6\sin(3x)\cos(3x) ).

So, ( \frac{d}{dx}[3\sin^2(3x)] = 6\sin(3x)\cos(3x) ).

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Answer from HIX Tutor

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