How do you differentiate #f(x) = (2 x^2 +5 x +10)^cos(x)#?

1 Answer
Jun 4, 2016

#f'(x)=((2x^2+5x+10)^cos(x)((4x+5)cos(x)-(2x^2+5x+10)sin(x)ln(2x^2+5x+10)))/(2x^2+5x+10)#

Explanation:

Let #y=(2x^2+5x+10)^cos(x)#.

Then

#ln(y)=ln((2x^2+5x+10)^cos(x))#

#ln(y)=cos(x)ln(2x^2+5x+10)#

Differentiating both sides, using the chain rule on the left and the product rule on the right:

#1/y(dy/dx)=-sin(x)ln(2x^2+5x+10)+cos(x)((4x+5)/(2x^2+5x+10))#

#1/y(dy/dx)=((4x+5)cos(x)-(2x^2+5x+10)sin(x)ln(2x^2+5x+10))/(2x^2+5x+10)#

Multiply both sides by #y#, or #(2x^2+5x+10)^cos(x)#,

#dy/dx=((2x^2+5x+10)^cos(x)((4x+5)cos(x)-(2x^2+5x+10)sin(x)ln(2x^2+5x+10)))/(2x^2+5x+10)#