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

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Answer 1 · 2016-06-01T04:53:28

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

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

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,
$3z^2sin ^2(2z)+4sin (2z)cos (2z)z^3$

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