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How do you differentiate #f(x)= (4-x^2) *ln x# using the product rule?

Answer 1

Isabella Ackerman

#((4-x^2)-2x^2 * lnx)/x#

Product rule: #h = f*g#

#h'= fg'+gf'#

Note: #f(x) = ln x#

#f'(x) = 1/x#

Given #f(x) = (4-x^2)*lnx#

#f'(x) = (4-x^2) d/dx(lnx) + lnx *d/dx(4-x^2)#

#= (4-x^2) (1/x) + -2x(lnx)#

# = (4-x^2)/x - (2x) (ln x) #

=#((4-x^2)-2x^2 * lnx)/x#

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Answer 2

Cameron Alexander

To differentiate the function ( f(x) = (4 - x^2) \cdot \ln(x) ) using the product rule, you apply the formula ( (uv)' = u'v + uv' ), where ( u = 4 - x^2 ) and ( v = \ln(x) ).

( u' = -2x ) (differentiating ( 4 - x^2 ) with respect to ( x ))

( v' = \frac{1}{x} ) (the derivative of ( \ln(x) ) is ( \frac{1}{x} ))

Now, applying the product rule:

( f'(x) = (4 - x^2) \cdot \frac{1}{x} + (\ln(x)) \cdot (-2x) )

Simplify:

( f'(x) = \frac{4 - x^2}{x} - 2x \ln(x) )

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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.