How do you differentiate #f(x)=(cos^3x)/sinx# using the quotient rule?

Question

Emma Aldridge · Accepted Answer

#f'(x)=(-cos^2x(3sin^2x+cos^2x))/sin^2x#.

Before finding the reqd. diifn, we find #d/dxcos^3x# by Chain Rule. #d/dxcos^3x=d/dx(cosx)^3=3cos^2x*d/dxcosx=-3cos^2xsinx#. Now, #f'(x)=(sinx*d/dxcos^3x-cos^3x*d/dxsinx)/sin^2x#...[Quotient Rule] #={(sinx)(-3cos^2xsinx)-(cos^3x)(cosx)}/sin^2x# #={-3cos^2xsin^2x-cos^4x}/sin^2x# #=(-cos^2x(3sin^2x+cos^2x))/sin^2x#.

Ezekiel Alexander · Answer

undefined To differentiate $ f(x) = \frac{\cos^3(x)}{\sin(x)} $ using the quotient rule, follow these steps:

1. Identify $ u $ and $ v $ in the function $ f(x) = \frac{u}{v} $.
2. Apply the quotient rule formula: $ f'(x) = \frac{u'v - uv'}{v^2} $.
3. Differentiate $ u = \cos^3(x) $ with respect to $ x $ to find $ u' $.
4. Differentiate $ v = \sin(x) $ with respect to $ x $ to find $ v' $.
5. Substitute $ u' $, $ v' $, $ u $, and $ v $ into the quotient rule formula.
6. Simplify the expression to obtain the derivative of $ f(x) $.

Following these steps will enable you to differentiate $ f(x) = \frac{\cos^3(x)}{\sin(x)} $ using the quotient rule.