1. ## Calulus help!

y= lnx/ 1+lnx find the derivative

integrate from 0 to pie over 2 tan (x/2) dx

integrate from pie over 2 to pie 2 cot x/3 dx

integrate dx/(2 sqaure root of x)(2x)

i have the answers in the back of the book but i have no idea how to solve! I really need to understand this.

also, suppose f(x) is positive, continuous, and increasing over the interval [a,b] by interpreting the graph of f show that:
integrate from a to b f(x)dx + integrate f(a) to f(b) f inverse(y) dy = bf(b)-af(a)

2. Originally Posted by genlovesmusic09
y= lnx/ 1+lnx find the derivative

use the quotient rule.

integrate from 0 to pie over 2 tan (x/2) dx

let $\textcolor{red}{u = \frac{x}{2}}$

$\textcolor{red}{dx = 2 \, du}$

$\textcolor{red}{2\int_0^{\frac{\pi}{4}} \frac{\sin{u}}{\cos{u}} \, du}$

integrate from pie over 2 to pie 2 cot x/3 dx

same method as above ... let $\textcolor{red}{u = \frac{x}{3}}$

integrate dx/(2 sqaure root of x)(2x)

note that $\textcolor{red}{2\sqrt{x} \cdot 2x = 4x^{\frac{3}{2}}}$
...

btw $\pi$ is spelled pi , not "pie"

3. wow i feel dumb (about the pi thing) typing to quickly...

for the first one i had tried quotient rule (before posting) and I get 1/x^2(l+lnx)^2

but the answer doesn't have an x^2 and i don't understand why...

4. Since you did try something, it would have been better to show what you had tried than just saying " i have no idea how to solve".

To differentiate $\frac{ln(x)}{1+ ln(x)}$, use, as you say, the quotient rule: $\frac{(ln(x))'(1+ ln(x))- (ln(x))(1+ ln(x))'}{(1+ ln(x))^2}$.

Now (ln(x))'= $\frac{1}{x}$ and (1+ ln(x))' is also $\frac{1}{x}$ since the derivative of "1" is 0. So the quotient rule gives $\frac{\frac{1}{x}(1+ ln(x))- \frac{1}{x}ln(x)}{(1+ln(x))^2}$
$= \frac{\frac{1}{x}}{(1+ ln(x))^2}= \frac{1}{x(1+ ln(x))^2}$

So I guess the question is where did you get the $x^2$ in the denominator?

5. $y = \frac{\ln{x}}{1+\ln{x}}$

$y' = \frac{(1+\ln{x}) \cdot \frac{1}{x} - \ln{x} \cdot \frac{1}{x}}{(1+\ln{x})^2}$

$y' = \frac{\frac{1}{x}[(1+\ln{x}) - \ln{x}]}{(1+\ln{x})^2}$

$y' = \frac{1}{x(1+\ln{x})^2}$

6. i found x^2 because I didn't take out 1/x out of both i left it and brought both x's to the bottem
now I see that i'm supposed to take it out thanks!