How do you differentiate #f(x)= (cos x)*e^-x*sqrt(3x)# using the product rule?

1 Answer
Nov 10, 2015

Answer:

#dy/dx = e^(-x)\*sqrt(3x)(cos(x)*(1-2x)/(2x)-sin(x))#

Explanation:

So we have

#y = cos(x)*e^(-x)*sqrt(3x)#

To simplify things, let's say #e^(-x)sqrt(3x) = u#

So we have

#y = u*cos(x)#

Using the product rule we have

#dy/dx = ud/dx(cos(x)) + cos(x)(du)/dx#
#dy/dx = -u*sin(x) + cos(x)(du)/dx#

And now we differentiate #u#

#u = e^(-x)* sqrt(3x) = e^(-x)*(sqrt(3)x^(1/2))#

#(du)/dx = e^(-x)d/dx(sqrt(3)x^(1/2)) + sqrt(3x)d/dx(e^(-x))#

#(du)/dx = e^(-x)*sqrt(3)/2x^(-1/2) -sqrt(3x)*e^(-x)#
#(du)/dx = e^(-x)((sqrt(3))/(2sqrt(x)) - sqrt(3x))#

Using some algebra to change that we have

#(du)/dx = e^(-x)((sqrt(3x))/(2x)-sqrt(3x))#
#(du)/dx = e^(-x)*sqrt(3x)(1/(2x) - 1)#
#(du)/dx = (u(1-2x))/(2x)#

Substitute that back in the original derivative

#dy/dx = -u*sin(x) + cos(x)*u\*(1-2x)/(2x)#

#dy/dx = u(cos(x)*(1-2x)/(2x) - sin(x))#

#dy/dx = e^(-x)\*sqrt(3x)(cos(x)*(1-2x)/(2x)-sin(x))#