Evaluating $\frac{\operatorname d \! \phantom x}{\operatorname d\!x}\frac{4}{\ln(x^2+2)}$

Question

$\dfrac{\operatorname d \! \phantom x}{\operatorname d\!x}\dfrac{4}{\ln(x^2+2)}= \dfrac{4}{\ln u}$

$u=x^2+2$
$u\;'=2x$

$y\;'=\dfrac{4}{\dfrac{1}{u}} \times (u\;') \implies \dfrac{4}{\dfrac{1}{x^2+2}}(2x) \implies (4x^2+8)(2x)=8x^3+16x$

What am I doing wrong?

Answer 1

Re-write the equation as $$y=4(\ln(x^2+2))^{-1}$$ Then use the chain rule and the power rule.

Let $u=x^2+2$, so $u'=2x$. Now you have $$y=4(\ln(u))^{-1}$$ This is just the power rule.

Let $v=\ln(u)$. Now you have $$y=4v^{-1}$$ $y'$ is just $$-1(v')4v^{-2}$$ $v'$ is just $$\frac{u'}{u}$$ which is $$\frac{2x}{x^2+2}$$ Put this all together and you have (answer is hidden under a "spoiler" - mouse over to see)

$$y'=\frac{-8x}{(x^2+2)(\ln(x^2+2))^2}$$

Answer 2

Your answer is wrong because when you differentiate with respect to u, you get:

$\frac{\mathrm{d} }{\mathrm{d} u}(\frac{4}{\ln u})=\frac{\frac{-4}{u}}{(\ln u)^2}=\frac{-4}{u(\ln u)^2}=\frac{-4}{(x^2+2)(\ln (x^2+2))^2}$

Applying the chain rule by multiplying with $\frac{\mathrm{du} }{\mathrm{d} x}=2x$, we get the answer of $\frac{-8x}{(x^2+2)(\ln (x^2+2))^2}$

Answer 3

. Put $v = \ln u$, and $u = x^2 + 2$.

$\dfrac{df}{dv} \left(\dfrac{4}{v}\right) = -\dfrac{4}{v^2}$,

$\dfrac{dv}{du} = \dfrac{1}{u}$

$\dfrac{du}{dx} = 2x$. Thus by the Chainrule:

$\dfrac{df}{dx} = \dfrac{df}{dv}\cdot \dfrac{dv}{du}\cdot \dfrac{du}{dx} = -\dfrac{4}{v^2}\cdot \dfrac{1}{u}\cdot 2x = \dfrac{-4}{\ln^2 (x^2+2)}\cdot \dfrac{1}{x^2+2}\cdot 2x = \dfrac{-8x}{(x^2+2)\ln^2 (x^2+2)}$