Question #56c3f

1 Answer
Nov 7, 2017

#(dy)/(dx)=-2cos(x^2+1)e^(-2x)(2xsin(x^2+1)+cos(x^2+1))#

Explanation:

#y=e^(-2x)cos^2(x^2+1)=uv#

If #y=f(x)g(x) => (dy)/(dx)=f(x)g'(x)+f'(x)g(x)#

#f(x)=e^(-2x)#
#f'(x)=-2e^(-2x)#

#g(x)=cos^2(x^2+1)=(cos(x^2+1))^2#
#g'(x)=2cos(x^2+1) * d/dx[cos(x^2+1)]#
#g'(x)=2cos(x^2+1) * -2xsin(x^2+1)#
#g'(x)=-4xsin(x^2+1)cos(x^2+1)#

#(dy)/(dx)=-4xsin(x^2+1)cos(x^2+1)e^(-2x)-2e^(-2x)cos^2(x^2+1)=-2cos(x^2+1)e^(-2x)(2xsin(x^2+1)+cos(x^2+1))#