Home > Calculus > Product Rule

How do you differentiate #f(x)=1/(x^3-1)x^2cosx# using the product rule?

Answer 1

Luke Alvarez

#f'(x)=-(x^4+2x)/(x^3-1)^2*cos(x)-x^2/(x^3-1)*sin(x)#

Using the product rule

#(uv)'=u'v+uv'# we get with #u=x^2/(x^3-1)#

#u'=(2x(x^3-1)-x^2*3x^2)/(x^3-1)^2# simplifying we get

#u'=-(x^4+2x)/(x^3-1)^2# #v=cos(x)# #v'=-sin(x)#

putting thins together we get

#f'(x)=-(x^4+2x)/(x^3-1)^2cos(x)-x^2/(x^3-1)sin(x)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Adam Adkins

To differentiate ( f(x) = \frac{x^2 \cos(x)}{x^3 - 1} ) using the product rule:

Let ( u(x) = x^2 ) and ( v(x) = \cos(x) )
Find ( u'(x) ) and ( v'(x) )
Apply the product rule:
( f'(x) = u'(x)v(x) + u(x)v'(x) )
Substitute the values into the formula:
( f'(x) = (2x)(\cos(x)) + (x^2)(-\sin(x)) )
Simplify the expression:
( f'(x) = 2x\cos(x) - x^2\sin(x) )

Therefore, the derivative of ( f(x) = \frac{x^2 \cos(x)}{x^3 - 1} ) using the product rule is ( f'(x) = 2x\cos(x) - x^2\sin(x) ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you differentiate #f(x)=1/(x^3-1)*x^2*cosx# using the product rule?

How do you differentiate #f(x)=1/(x^3-1)x^2cosx# using the product rule?