Home > Calculus > Product Rule

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

Answer 1

Isaiah Allen

#f'(x)=6x^2-x^-2-2x^-3-2#

The product rule is # d/dx(uv)=u (dv)/dx+v (du)/dx #; so #f'(x)=(x^3+2x+1)d/dx(2+x^-2)+d/dx(x^3+2x+1)(2+x^-2)#

#f'(x)=(x^3+2x+1)(-2x^-3)+(3x^2+2)(2+x^-2)#

#f'(x)=(-2)(x^3+2x+1)/(x^3)+(6x^2+3x^-2+4+2x^-2)#

#f'(x)=(-2)(1+2x^-2+x^-3)+(6x^2+4+5x^-2)#

#f'(x)=-2-4x^-2-2x^-3+6x^2+4+5x^-2#

#f'(x)=6x^2-x^-2-2x^-3-2#

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

Answer 2

Emma Aldridge

To differentiate ( f(x) = (x^3 + 2x + 1)(2 + x^{-2}) ) using the product rule, you apply the formula:

[ \frac{d}{dx}(uv) = u'\cdot v + u \cdot v' ]

Where ( u = x^3 + 2x + 1 ) and ( v = 2 + x^{-2} ).

The derivatives are:

[ u' = 3x^2 + 2 ] [ v' = -2x^{-3} ]

Now, applying the product rule:

[ f'(x) = (3x^2 + 2)(2 + x^{-2}) + (x^3 + 2x + 1)(-2x^{-3}) ]

[ f'(x) = (6x^2 + 4) + \frac{3x^3}{x^2} - \frac{4x}{x^2} - \frac{2}{x^2} ]

[ f'(x) = 6x^2 + 4 + 3x - 4x^{-1} - 2x^{-2} ]

So, the derivative of ( f(x) ) is ( f'(x) = 6x^2 + 3x + 4 - \frac{4}{x} - \frac{2}{x^2} ).

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.