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How do you differentiate #g(z) = z^3sin^2(2z)# using the product rule?

Answer 1

Evelyn Agard

#frac{d}{dz}(z^3sin ^2(2z))=3z^2sin ^2(2z)+4sin (2z)cos (2z)z^3#

#\frac{d}{dz}(z^3sin ^2(2z))#

Applying product rule, #(fcdot g)^'=f^'cdot g+fcdot g^'#

#f=z^3# ; #g=sin^2(2z)#

#=frac{d}{dz}(z^3)sin ^2(2z)+frac{d}{dz}(sin ^2(2z))z^3#

We know, #frac{d}{dz}(z^3)=3z^2# also, #frac{d}{dz}(sin ^2(2z))=4sin (2z)cos (2z)#

So finally,we get, #3z^2sin ^2(2z)+4sin (2z)cos (2z)z^3#

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Answer 2

Luke Alvarez

To differentiate ( g(z) = z^3\sin^2(2z) ) using the product rule:

Identify the two functions being multiplied: ( f(z) = z^3 ) and ( h(z) = \sin^2(2z) ).
Apply the product rule: ( g'(z) = f'(z)h(z) + f(z)h'(z) ).
Find the derivatives of ( f(z) ) and ( h(z) ):
- ( f'(z) = 3z^2 ) (using the power rule)
- ( h'(z) = 2\sin(2z)\cos(2z) ) (using the chain rule)
Substitute these derivatives into the product rule formula: ( g'(z) = 3z^2 \sin^2(2z) + z^3(2\sin(2z)\cos(2z)) ).
Simplify the expression if necessary, which may involve applying trigonometric identities.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.