Find the Derivatives of $g(x) = \sqrt{3-2x^2}$ and $h(x) = \ln {(x^2 – x)}$

Question

I am asked to find the derivatives of $g(x) =\sqrt{3-2x^2}$ and $h(x) = \ln{(x^2 – x)} $

For:

$g(x)h(x)$

and

$\dfrac{h(x)}{g(x)}$

and

$h^3 (x)$

First off I am not sure if my derivatives are correct. Here is what I have..

$g'(x) = \dfrac{2x} { \sqrt{3 - 2x^2}}$

$h'(x) =\dfrac {2x - 1} {x^2 - x}$

Thanks Em

Do you mean that $g(x)=\sqrt{3-2x^2}$ and $h(x)=\ln(x^2-x)$? — Kenny Lau, Jun 25 '16 at 05:49
$g'(x)$ is missing a negative sign. Other than that, your derivatives are correct. — Kenny Lau, Jun 25 '16 at 05:50
Hi yes that is what I mean. Sorry but I do not know what LATEX is but I will try and find out, — user350037, Jun 25 '16 at 05:59
Hi Kenny. I have had a look through the question and I do not see why g'(x) needs a negative sign. The 4x is being multiplied by + 0.5 not negative. Thoughts? — user350037, Jun 25 '16 at 16:17

score 0 · Accepted Answer · answered Jun 25 '16 at 06:13

Your $h'$ is correct, but your $g'$ should be $\dfrac 12 \cdot \dfrac{-4x}{\sqrt{3-2x^2}}=\dfrac {-2x}{\sqrt{3-2x^2}}$

So from the product rule we have $(h \cdot g)'=h' \cdot g + h \cdot g'=\dfrac {2x-1}{x^2-1} \cdot \sqrt{3-2x^2} + \ln (x^2-x) \cdot \dfrac {-2x}{\sqrt{3-2x^2}}$. You can't really do much to simplify it, except for making the whole second tern negative instead of having $-2x$ in the numerator.

$\dfrac hg$ is exactly the same, except a different formula.

For $h^3$, we use the chain rule: $(h^3)'=h' \cdot 3(h)^2=\dfrac {2x-1}{x^2-1} \cdot 3(\ln (x^2-x))^2$. Can't do much to simplify this either.

Find the Derivatives of $g(x) = \sqrt{3-2x^2}$ and $h(x) = \ln {(x^2 – x)}$

1 Answers1