How do you find the derivative of #ln (x^8)/ x^2#?

Answer 1

In this case, you would use quotient rule.

Before we get started, check this out. Here is the definition for quotient rule.

#f(x)=g(x)/(h(x))#
#f'(x)=(h(x)g'(x) - h'(x)g(x))/(h(x))^2#
So know what we know the definition, let's apply it. We know that #g(x)=ln(x^8)# and #h(x)=x^2#. So now, it's just plug n' solve.
#f'(x)=(x^2((8x^7)/(x^8))-2x(ln(x^8)))/(x^4)#

And I would personally be satisfied with the below answer.

#f'(x)=(8x-2xln(x^8))/(x^4)#
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Answer 2

To find the derivative of (\frac{{\ln(x^8)}}{{x^2}}), use the quotient rule:

Let (u = \ln(x^8)) and (v = x^2).

Apply the quotient rule: [ \frac{{d}}{{dx}}\left(\frac{{\ln(x^8)}}{{x^2}}\right) = \frac{{v \frac{{du}}{{dx}} - u \frac{{dv}}{{dx}}}}{{v^2}}]

Now, find (\frac{{du}}{{dx}}) and (\frac{{dv}}{{dx}}): [ \frac{{du}}{{dx}} = \frac{{8}}{{x}}] [ \frac{{dv}}{{dx}} = 2x]

Substitute these derivatives into the quotient rule: [ \frac{{d}}{{dx}}\left(\frac{{\ln(x^8)}}{{x^2}}\right) = \frac{{x^2 \cdot \frac{{8}}{{x}} - \ln(x^8) \cdot 2x}}{{(x^2)^2}}]

Simplify: [ \frac{{d}}{{dx}}\left(\frac{{\ln(x^8)}}{{x^2}}\right) = \frac{{8 - 2x\ln(x^8)}}{{x^4}}]

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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.

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.

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.

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.

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