# Thread: one derivative and one integration

1. ## one derivative and one integration

I am having trouble with this derivative:
y = (2x+3/2x-3)^(1/4)
The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
ln(y) = ln((2x+3/2x-3)^(1/4))

The other problem I am having is this integration:
(Integral sign) (xdx)/(x^4+3) The x is raised to just 4. the 3 is not an exponent.
I substituted u to be equal to x^2 and then du = 2xdx.
I now am stuck with 1/2(Integral sign) (du)/(u^2+3) The u is raised to 2..the 3 is not an exponent.

Thanks for all the help!

2. Originally Posted by davecs77
y = (2x+3/2x-3)^(1/4)
The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
ln(y) = ln((2x+3/2x-3)^(1/4))
You can do it by taking logarithms, but you don't need to. It's just a chain rule:
$y = f(g(x))$ so $y^{\prime} = f^{\prime}(g(x)) \cdot g^{\prime}(x)$

$y = \left ( \frac{2x + 3}{2x - 3} \right )^{1/4}$

$y^{\prime} = \frac{1}{4} \left ( \frac{2x + 3}{2x - 3} \right )^{-3/4} \cdot \frac{2(2x - 3) - (2x + 3)(2)}{(2x - 3)^2}$

$y^{\prime} = -3 \left ( \frac{(2x + 3)^{-3/4}}{(2x - 3)^{5/4}} \right )$

-Dan

3. Originally Posted by davecs77
y = (2x+3/2x-3)^(1/4)
The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
ln(y) = ln((2x+3/2x-3)^(1/4))
Doing it using logarithms,
$ln(y) = ln \left ( \left ( \frac{2x + 3}{2x - 3} \right )^{1/4} \right ) = \frac{1}{4} ln \left ( \frac{2x + 3}{2x - 3} \right ) = \frac{1}{4} ln(2x + 3) - \frac{1}{4} ln(2x - 3)$

Now take the derivative of both sides:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{4} \frac{1}{2x + 3} \cdot 2 - \frac{1}{4} \frac{1}{2x - 3} \cdot 2$

$\frac{dy}{dx} = y \cdot \left ( \frac{1}{2(2x + 3)} - \frac{1}{2(2x - 3)} \right )$

Now sub in your given value of y and simplify. You should (if I did it right) get the same expression I got in the first post.

-Dan

4. Originally Posted by davecs77
The other problem I am having is this integration:
(Integral sign) (xdx)/(x^4+3) The x is raised to just 4. the 3 is not an exponent.
I substituted u to be equal to x^2 and then du = 2xdx.
I now am stuck with 1/2(Integral sign) (du)/(u^2+3) The u is raised to 2..the 3 is not an exponent.
Look at
$\int \frac{du}{u^2 + 3}$

Substitute $u = \sqrt{3}z$. Then
$\int \frac{du}{u^2 + 3} = \sqrt{3} \int \frac{dz}{3z^2 + 3} = \frac{\sqrt{3}}{3} \int \frac{dz}{z^2 + 1}$

Does this integral look more familiar? (Think trig substitution...)

-Dan

5. Originally Posted by topsquark
Look at
$\int \frac{du}{u^2 + 3}$
$\int\frac1{a^2+u^2}~du=\frac1{a}\arctan\frac{u}{a} +k,~~a\ne0$