Home > Calculus > Differentiating Logarithmic Functions with Base e

Hoe do you differentiate #f(x)=xln(1/x^2) #?

Answer 1

Use some algebra to change the form of the function, and then apply the product rule to find
#f'(x)= ln(1/x^2)-2#

If we were to differentiate directly, we would need to use the product rule, the chain rule, and the quotient rule. However, with some basic algebra, we can make it so all we need is the product rule.

First, the algebra:

#f(x) = xln(1/(x^2)) = xln(x^(-2)) = -2xln(x)#

Now, given #d/dx-2x = -2# and #d/dx ln(x) = 1/x#

we use the product rule to obtain

#f'(x) = d/dx(-2x)ln(x)#

# = (-2x)(d/dxln(x)) + (d/dx-2x)(ln(x))#

#= -2x(1/x) + (-2)ln(x)#

#= -2ln(x) - 2#

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

#= ln(1/x^2)-2#

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Answer 2

Elijah Abram

To differentiate the function ( f(x) = x \ln \left( \frac{1}{x^2} \right) ), you can use the product rule.

Let ( u = x ) and ( v = \ln \left( \frac{1}{x^2} \right) ). Then, ( u' = 1 ) and ( v' = \frac{-2}{x} ).

Using the product rule, the derivative of ( f(x) ) is:

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

[ f'(x) = (1)\ln \left( \frac{1}{x^2} \right) + (x)\left( \frac{-2}{x} \right) ]

[ f'(x) = \ln \left( \frac{1}{x^2} \right) - \frac{2}{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.