To find the first derivative #dy/dx# we must apply the chain rule which states:
#dy/dx[f'(g(x))]=f'(g(x))*g'(x)#
Let
#f(x)=sin(x)#
#g(x)=e^-x#
And so,
#f'(x)=cos(x)#
#g'(x)=e^-x*d/dx(-x)=e^-x*-1=-e^-x#
Thus:
#dy/dx[sin(e^-x)]=cos(e^-x)*-e^-x#
We can simplify this and rewrite as:
#color(red)(dy/dx[sin(e^-x)]=-e^-xcos(e^-x)#
To find the second derivative we must now use the product rule given that we have a product of two functions:
The product rule states: #dy/dx[(f*g)(x)]=f'(x)*g(x)+f(x)*g'(x)#
#f(x)=-e^-x#
#g(x)=cos(e^-x)#
And so,
#f'(x)=-e^-x*d/dx(-x)=-e^-x*-1=e^-x#
#g'(x)=-sin(e^-x)*d/dx(e^-x)=-sin(e^-x)*-e^-x=e^-xsin(e^-x)#
Thus:
#(d^2y)/dx^2=(e^-x)*cos(e^-x)+(-e^-x)(e^-xsin(e^-x))#
Simplifying we get
#color(red)((d^2y)/dx^2=e^-xcos(e^-x)-e^(-2x)sin(e^-x)#
