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

Question

Scarlett Allison · Accepted Answer

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

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)#

Elijah Abram · Answer

undefined To differentiate the given function $ y = \frac{\sqrt{\cos x}}{5\ln x} $, you would use the quotient rule of differentiation:

$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} $$

Where $ u = \sqrt{\cos x} $ and $ v = 5\ln x $.

Now, differentiate $ u $ and $ v $ separately:

$$ u' = \frac{-\sin x}{2\sqrt{\cos x}} $$
$$ v' = \frac{5}{x} $$

Now apply the quotient rule:

$$ \frac{d}{dx}\left(\frac{\sqrt{\cos x}}{5\ln x}\right) = \frac{\frac{-\sin x}{2\sqrt{\cos x}} \cdot (5\ln x) - \sqrt{\cos x} \cdot \frac{5}{x}}{(5\ln x)^2} $$

Simplify the expression:

$$ \frac{-5\sin x \ln x}{2x\sqrt{\cos x}} - \frac{5\sqrt{\cos x}}{x(5\ln x)^2} $$

This is the derivative of the given function.