How do you use the chain rule to differentiate #f(x)=cos(2x^2+3x-sinx)#?

Question

1199 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-27T11:55:55

#f'(x)=-(4x+3-cosx)sin(2x^2+3x-sinx)#

#(dy)/(dx)=color(red)((dy)/(du))xxcolor(blue)((du)/(dx)#

let#" "y=f(x)=cos(2x^2+3x-sinx)#

#color(blue)(u=2x^2+3x-sinx=>(du)/(dx)=4x+3-cosx)#

#:.color(red)(y=cosu=>(dy)/(du)=-sinu).#

#f'(x)=(dy)/(dx)=color(red)((-sinu))xxcolor(blue)((4x+3-cosx)#

#f'(x)=-(4x+3-cosx)sin(2x^2+3x-sinx)#