$\frac{d}{d\theta}\mu=?$, where $\theta=\log\frac{\mu}{1-\mu}$

Question

How can I solve the following differentiation :

$$\frac{d}{d\theta}\mu,$$

where $\theta=\log\frac{\mu}{1-\mu}$ ?

Answer 1

As $\dfrac{dy}{dx}\cdot\dfrac{dx}{dy}=1,$ See Is $ \frac{\mathrm{d}{x}}{\mathrm{d}{y}} = \frac{1}{\left( \frac{\mathrm{d}{y}}{\mathrm{d}{x}} \right)} $?

$\dfrac{d\mu}{d\theta}=\dfrac1{\dfrac{d\theta}{d\mu}}$

Now $\dfrac{d\theta}{d\mu}=\dfrac{d(\ln\mu)}{d\mu}-\dfrac{d\{\ln(1-\mu)\}}{d\mu}=?$

Answer 2

You will need the expression of $\mu$ in terms of $\theta$, when what you have is $\theta$ in terms of $\mu$

Since $$ \theta = \log{\frac{\mu}{1 - \mu}} \\ e^\theta = \frac{\mu}{1 - \mu} \\ e^\theta(1 - \mu) = \mu \\ e^\theta = \mu(1 + e^\theta) \\ \mu = \frac{e^\theta}{1 + e^\theta} $$

Now, differentiate $\mu$ using the chain rule, that

$$ \frac{d\mu}{d\theta} = \frac{d}{d\theta}\frac{e^\theta}{1 + e^\theta} = \\ e^\theta \frac{d}{d\theta}\frac{1}{1 + e^\theta} + \frac{1}{1 + e^\theta}\frac{d e^\theta}{d\theta} $$

Solve the final expression for the answer you want

Answer 3

The way I would do it is as such: $$e^\theta = \frac{\mu}{1-\mu} \implies \frac{e^\theta}{1+e^\theta}=\mu$$ You can differentiate this as you wish; using the quotient rule we get $$\frac{d}{d\theta} \frac{e^\theta}{1+e^\theta} = \frac{e^\theta(1+e^\theta) - e^{2\theta}}{(1+e^\theta)^2}$$ Feel free to clean up that last expression as you like