What is the derivative of #(cosx)^x#?

2 Answers
May 28, 2017

Answer:

#d/dx((cosx)^x) = (lncosx -x/sinx)(cosx)^x#

Explanation:

Let #f(x) =(cosx)^x#

#lnf(x)= xlncosx#

#(f'(x))/(f(x)) = lncosx - x/sinx#

#f'(x) = f(x)(lncosx-x/sinx)#

#f'(x) = (lncosx -x/sinx)(cosx)^x#

May 28, 2017

Answer:

#d/dxcos^x(x)=cos^x(x)(-xtan(x)+ln(cos(x)))#

Explanation:

Step 1. Express #cos(x)^x# as a power of #e#.

#d/dxcos^x(x)=d/dxe^(ln(cos^x(x)))=d/dxe^(xln(cos(x)))#

Step 2. Use the chain rule with #u=xln(cos(x))# and #d/(du)(e^u)=e^u#

Chain Rule: #d/dx(e^(xln(cos(x))))=(de^u)/(du)(du)/(dx)#

#(de^u)/(du)=e^u#

#(du)/(dx)=d/dxxln(cos(x))#

Using the Product Rule

#(du)/(dx)=d/dxxln(cos(x))=x d/dxln(cos(x))+ln(cos(x))dx/dx#

#(du)/(dx)=x(1/cos(x)(-sin(x)))+ln(cos(x))(1)#

#(du)/(dx)=-xtan(x)+ln(cos(x))#

Plugging it back in to the Chain Rule

#color(blue)((de^u)/(du))color(red)((du)/(dx))=color(blue)(e^(xln(cos(x))))color(red)((-xtan(x)+ln(cos(x))))#

Simplifying a bit gives

#(de^u)/(du)(du)/(dx)=cos^x(x)(-xtan(x)+ln(cos(x)))#