0
$\begingroup$

I have the function:

$\frac{x-1}{x^3}$

I differentiated this using the quotient rule and got:

$-\frac{2x-3}{x^4}$

What I need to do and can't seem to work out is how to differentiate the same function by (a) chain rule and then (b) "applying the properties of logarithms to break up the expression into simpler logarithms first".

$\endgroup$
2
  • 1
    $\begingroup$ I see no logarithms in the original function. The question would make a lot more sense if the given function was $\log \frac{x-1}{x^3}$. $\endgroup$ – Florian Jun 19 '19 at 8:32
  • 2
    $\begingroup$ @Florian. Logarithmic differentiation replaces the chaine rule, the quotient rule and the power law. $\endgroup$ – Claude Leibovici Jun 19 '19 at 8:34
3
$\begingroup$

Let $$y=\frac{x-1}{x^3}$$

Then $$\log y=\log (x-1)- \log (x^3)$$

$$\Rightarrow \log y=\log (x-1)- 3\log (x)$$

$$\Rightarrow \frac 1y\frac{dy}{dx}=\frac 1{x-1}-\frac 3x$$

$$\Rightarrow \frac{x^3}{x-1}\frac{dy}{dx}=\frac 1{x-1}-\frac 3x$$

$$\Rightarrow \frac{dy}{dx}=\frac 1{x^3}-\frac {3(x-1)}{x^4}$$

$$\Rightarrow \frac{dy}{dx}=\frac x{x^4}-\frac {3x-3}{x^4}$$

$$\Rightarrow \frac{dy}{dx}=-\frac {2x-3}{x^4}$$ which is the same as you have found.

By chain rule I expect they mean for you to recognise that $$y=\frac{x-1}{x^3}=\frac{1}{x^2}-\frac{1}{x^3}=(x)^{-2}-(x)^{-3}$$

$$\Rightarrow \frac{dy}{dx}=(-2)(x)^{-3}-(-3)(x)^{-4}$$

$$\Rightarrow \frac{dy}{dx}=-\frac {2}{x^3}+\frac {3}{x^4}=-\frac {2x-3}{x^4}$$ which is the same again.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.