How do you find the derivative of #f(x) = (cos x) e^(-xsqrt3)#?

Question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-04T15:45:32

We can use the product rule, which states that, be #y=f(x)g(x)#, then

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

So, we need the elements:

#f(x)=cos(x)#
#f'(x)=-sen(x)#
#g(x)=e^(-x*sqrt(3))#
#g'(x)# demands chain rule. The chain rule states that the derivative is given by #(dy)/(dx)=(dy)/(du)(du)/(dx)#. Thus, renaming #u=-xsqrt(3)#, we have that:

#g'(x)=-sqrt(3)e^(-xsqrt(3))#

Solving the original derivation, then:

#(dy)/(dx)=-sen(x)e^(-xsqrt(3))+cos(x)(-sqrt(3)e^(-xsqrt(3)))#

As stated by a law of exponentials, #a^-n=1/a^n#

#(dy)/(dx)=-(sen(x))/e^(xsqrt(3))-(sqrt(3)cos(x))/e^(xsqrt(3))#

#(dy)/(dx)=-(sen(x)+sqrt(3)cos(x))/e^(xsqrt(3))#