How do you differentiate $g(x) = sqrt(2x^2-1)cos3x$ using the product rule?

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Answer 1 · 2016-05-04T15:19:35

$g' (x)=-3*sin (3x)sqrt(2x^2-1)+ (2x*cos 3x)/(sqrt(2x^2-1))$

The formula for the product rule is

$d/dx(uv)=u dv/dx+ v*du/dx$

from the given $g(x)=sqrt(2x^2-1)*cos 3x$

We let $u=sqrt(2x^2-1)$ and $v=cos 3x$

$d/dx(uv)=u dv/dx+ v*du/dx$

$g' (x)=d/dx(sqrt(2x^2-1)*cos 3x)=(sqrt(2x^2-1)) d/dx(cos 3x)+ (cos 3x)*d/dxsqrt(2x^2-1)$

$g' (x)=$
$(sqrt(2x^2-1))*(-3 sin 3x)+ (cos 3x)*1/(2*sqrt(2x^2-1))*(4x-0)$

$g' (x)=-3*sin (3x)sqrt(2x^2-1)+ (2x*cos 3x)/(sqrt(2x^2-1))$

God bless....I hope the explanation is useful.

How do you differentiate g(x) = sqrt(2x^2-1)cos3xg(x) = sqrt(2x^2-1)cos3x using the product rule?