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

Question

1348 views around the world

You can reuse this answer
Creative Commons License

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.