How do you differentiate #y = (sqrt(cos x))/(5lnx)#?

1 Answer
May 22, 2018

Answer:

#-(xln(x)sin(x)+2cos(x))/(10xsqrt(cos(x))(ln(x))^2)#

Explanation:

use quotient rule:

#d/dx(f(x)/g(x))=(f'(x)g(x)-g'(x)f(x))/(g(x))^2#

(see this proof or this)

using above format for this question, #f(x)=sqrt(cosx)# and #g(x)=5ln(x)#

#y'(x)=(d/dx(sqrt(cos(x)))(5ln(x))-d/dx(5ln(x))sqrt(cos(x)))/(5ln(x))^2#

#y'(x)=(1/(2sqrt(cos(x)))*d/dx(cos(x))(5ln(x))-(5/x)sqrt(cos(x)))/(5ln(x))^2#

#y'(x)=(1/(2sqrt(cos(x)))*(-sin(x))(5ln(x))-(5/x)sqrt(cos(x)))/(5ln(x))^2#

when you simplify you should get something like:

#-(xln(x)sin(x)+2cos(x))/(10xsqrt(cos(x))(ln(x))^2)#