Evaluating $\frac{d}{dx}\sqrt[4]{\ln(12-x^2)}$

Question

Find Derivative and evaluate at $x=1$:

$$ \frac{d}{dx}\sqrt[4]{\ln(12-x^2)} = (\ln u)^{1/4} $$

$$v=(v)^{1/4} \implies v=\ln\;u, v\;'=\dfrac{1}{u}(u\;')$$

$$y\;'=\frac{1}{4}v^{-3/4}\; \times v\;'= \dfrac{1}{4}\dfrac{1}{v^{3/4}}v\;'$$

$$\frac{1}{4} \dfrac{1}{\ln(12-x^2)^{3/4}} \dfrac{1}{\ln(12-x^2)} (-2x)$$

Is this correct? It doesn't look correct. I evaluated at $1$ and got $-0.10821.$

Answer 1

$$\big (\sqrt[4]{\ln(12-x^2)}\big)'\big|_{x=1} =\frac 14(\ln(12-x^2))^{-\frac34}\frac{1}{12-x^2}(-2x)\big|_{x=1}=-\frac{(\ln 11)^{-\frac34}}{22}.$$

Answer 2

You did not calculate the derivative of $\ln(12-x^2)$ correctly.

Answer 3

It is not correct, but you are really close: you miscalculated the derivative of $\log(12 - x^2)$, which should be just $\frac{1}{12 - x^2} \cdot (-2x)$.

Answer 4

The secret: Applying the chain rule until it stops hurting.

Reminder: $(f(g(x)))' =g'(x)f'(g(x)) $. Also $(\ln f(x))' =\frac{f'(x)}{f(x)} $ and $(f(x)^a)' =f'(x)f(x)^{a-1} $

$\begin{array}\\ \dfrac{d}{dx}\sqrt[4]{\ln(12-x^2)} &=((\ln(12-x^2))^{1/4})'\\ &=(\ln(12-x^2))'(\ln(12-x^2))^{-3/4}\\ &=\dfrac{(12-x^2)'}{12-x^2}(\ln(12-x^2))^{-3/4}\\ &=\dfrac{-2x}{12-x^2}(\ln(12-x^2))^{-3/4}\\ \end{array} $