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How do you differentiate #f(x)= (4x^2-1) *ln x# using the product rule?

Answer 1

Emilia Aguilar

#y'=(4x^2-1)*1/x+lnx*8x#

The product Rule:

#color(red)[y=f(x)*g(x)]#

#color(red)[y'=f(x)*g'(x)+g(x)*f'(x)]#

now lets derive #color(blue)[y= (4x^2-1) *ln x]#

#y'=(4x^2-1)*1/x+lnx*8x#

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Answer 2

Ezekiel Alexander

To differentiate ( f(x) = (4x^2 - 1) \cdot \ln x ) using the product rule, apply the formula:

[ (uv)' = u'v + uv' ]

where ( u = 4x^2 - 1 ) and ( v = \ln x ). Then, differentiate each term:

[ u' = 8x ] [ v' = \frac{1}{x} ]

Finally, substitute these derivatives into the product rule formula:

[ f'(x) = (4x^2 - 1) \cdot \frac{1}{x} + (8x) \cdot \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.