Explanation for differentiating $y= \frac 1{\ln x}$

Question

The solution in my book shows

$$\frac{dy}{dx} = \frac{\ln(x) \cdot 0 - 1 \cdot \frac{1}{x}}{\ln^2(x)}$$

Can you please explain how the differentiation is done?

I can easily differentiate it with substitution and I can understand the denominator easily but I have no idea how he came up with that numerator

Answer 1

y = 1/ln x

y= log e (base x)

x^y = e

taking log of both sides

y ln x =ln e

differentiate wrt x

y(1/x)+y'(ln x) = 0

y'(ln x) = -(y/x)

y' =-y/ x(ln x)

y' =-1/x(ln x)^2

Answer 2

$y=\frac{1}{\ln(x)}=(\ln(x))^{-1}.$

Now, use the power rule for differentiation and then apply the chain rule.

$y'=(-1)(\ln(x))^{-2}\frac{1}{x}=-\frac{1}{x\ln(x)^2}$.

An Alternative way would be to use the quotient rule:

$y=\frac{1}{\ln(x)}=\frac{u}{v} \implies y'=\frac{u'v-uv'}{v^2}=\frac{(0)'\ln(x)-1(\ln(x))'}{\ln(x)^2}=-\frac{1/x}{\ln(x)^2}$

Answer 3

This is quotient rule of differentiation.

$$\frac{dy}{dx}\left(\frac{f(x)}{g(x)}\right) =\frac{f'(x) \cdot g(x)-g'(x)\cdot f(x)}{\big(g(x)\big)^2}$$

Answer 4

Your book's solution is using the quotient rule, which says that if $f(x) = \frac{u(x)}{v(x)}$, then

$$f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$$

In this case, $u(x) = 1$ and $v(x) = \ln(x)$.

Answer 5

The books solution clearly makes use of the quotient rule. But when you differentiate a quotient with numerator a constant (like in this case, in which the numerator is equal to $1$, it can often be quicker to differentiate using the chain rule:

Recall the chain rule:

Suppose $y = g(f(x))$. Then $y' = \dfrac{dy}{dx} = g'(f(x)) \cdot f'(x).\;$ We can use it here, too, designating $f(x) = \ln x$, and $g(x) = x^{-1}.$

And hence $\;y= g(f(x)) = (\ln x)^{-1}$.

Using the the chain rule, we get : $$\frac{dy}{dx} = \underbrace{- \ln(x)^{-2}}_{\large g'(f(x))} \cdot \underbrace{\frac 1x}_{\large f'(x)} =\; \color{blue}{\bf -\frac 1{(x\ln(x)^2)}}$$