# Thread: More differentiation... blah

1. ## More differentiation... blah

I hate to post a bunch of questions... just cuz I'm lost... but I guess I'll post the questions and what I got and whoever can help me out with whatever they can help me out with would be nice

I was supposed to solve these by logarithmic differentiation...

y= (lnx)^(2x)

i got dy/dx = (lnx)^(2x) * 2((1+lnx*ln(lnx))/lnx)

this next one was a big mess so I don't know if anyone would want to take it on... no need to simplify though

logarithmic differentiation again...

y = (x^(1/2) * e^(x)^(5) * (x^(2) + 1)^(1/3))/(x+1)

i got... a mess... I don't even think I can type it all out... blah... if this one is too much trouble I understand =)

this one seemed simple enough just wanted to make sure I was doing things correctly

x^(y) = y^(x)

i got dy/dx = (-y^(2) + xylny)/(xylnx - x^(2))

These next 2 are just regular differentiation... any help is appreciated.

h(x) = ln(x + (x^(2) - 1)^(1/2))

i got

h'(x) = 1/(x(x^(2))^(1/2) + x - 1)

last one i promise =)

f(x) = ln [((2x^(2) + 1)/(2x-1))^(1/2)]

i came up with

(2x^(2) - 2x - 1)/[(2x^(2)+1) * (2x-1)]

THANK YOU SO MUCH for any help... I really do appreciate it greatly.

2. Hello, drain!

I'll do the last two . . . and thank you for showing your work!

h(x) .= .ln[x + (x² - 1)^½]

. . . . . . . 1 + ½(x² - 1)^{-½}·2x . . . . 1 + x(x² - 1)^{-½}
h'(x) . = . ---------------------------- . = . -----------------------
. . . . . . . . . x + (x² - 1)^½ . . . . . . . . . .x + (x² - 1)^½

Multiply top and bottom by (x² - 1)^½

. . . . . . . . . (x² - 1)^½ + x
. . = . --------------------------------- . . . . This reduces!
. . . . .(x² - 1)^½
[x + (x² - 1)^½]

. . . . . . . . . . . . 1
. . h'(x) .= .--------------
. . . . . . . . .(x² - 1)^½

f(x) .= .ln[(2x² + 1)/(2x - 1)^½]

We have: .f(x) .= .ln(2x² + 1) - ½·ln(2x - 1)

. . . . . . . . . . . . . . 4x . . . . . . . 2 . . . . . . .6x² - 4x - 1
Then: . f'(x) . = . ---------- - ½·------- . = . --------------------
. . . . . . . . . . . . .2x² + 1 . . . 2x - 1 . . . .(2x² + 1)(2x - 1)

3. thanks for your help man! But on that last one you did, I don't know if I mistyped it, or you misread it, but the whole thing was supposed to be square rooted. And sorry for not showing my work... if you're going to take the time to help me... I guess I should take the time to help you help me

It was supposed to be ln[sqrt{(2x^(2)+1)/(2x-1)}]

First I got

1/sqrt[(2x^(2) + 1)/(2x-1)] * 1/2 * [(2x^(2) + 1)/(2x-1)]^(-1/2) * [(2x-1)(4x) - (2x^(2) + 1)(2)]/(2x-1)^2] by using the quotien rule

then

(4x^(2) - 4x - 2)/2(2x^(2) + 1)(2x-1) which i factored out a 2 and got

(2x^(2) - 2x - 1)/[(2x^(2) * (2x-1)]

4. Originally Posted by drain
thanks for your help man! But on that last one you did, I don't know if I mistyped it, or you misread it, but the whole thing was supposed to be square rooted. And sorry for not showing my work... if you're going to take the time to help me... I guess I should take the time to help you help me

It was supposed to be ln[sqrt{(2x^(2)+1)/(2x-1)}]

First I got

1/sqrt[(2x^(2) + 1)/(2x-1)] * 1/2 * [(2x^(2) + 1)/(2x-1)]^(-1/2) * [(2x-1)(4x) - (2x^(2) + 1)(2)]/(2x-1)^2] by using the quotien rule

then

(4x^(2) - 4x - 2)/2(2x^(2) + 1)(2x-1) which i factored out a 2 and got

(2x^(2) - 2x - 1)/[(2x^(2) * (2x-1)]
Soroban's way of simplifying the ln before differentiating is still best here. It saves you from having to use the bothersome quotient rule.

f(x) = ln [((2x^(2) + 1)/(2x-1))^(1/2)]
=> f(x) = (1/2)*[ ln(2x^(2) + 1) - ln(2x-1)]
=> f ' (x) = (1/2) * [1/(2x^(2) + 1) * 4x - 1/(2x - 1) * 2]
=> f ' (x) = (1/2) * [4x/(2x^(2) + 1) - 2/(2x - 1)]
=> f ' (x) = 2x/(2x^(2) + 1) - 1/(2x - 1)
=> f ' (x) = (4x^2 - 2x - 2x^2 - 1)/[(2x^(2) + 1)(2x - 1)]
=> f ' (x) = (2x^(2) - 2x - 1)/[(2x^(2)+1) * (2x-1)]
which is what you got--but with slightly harder work. Sure, it looks like i took more steps, but most were easy simplifications and could be omitted

5. Yeah I didn't notice that you could factor out the exponent, until I was re-doing it Thanks to both of you!!!

6. Hello again, drain!

Yes, I misplaced the square root . . . my fault.
. . Luckily, Jhevon covered for me.

x^y .= .y^x

I got: .dy/dx .= .[-y² + xy·ln(y]/[xy·ln(x) - x²] . . . . Yes!
y . = . x^½ · e^{x^5) · [(x² + 1)^{1/3}]/(x + 1)
This one is messy . . . I won't even try to simplify it.

Take logs: . ln(y) . = . (1/2).ln(x) + x^5 + (1/3)·ln(x² + 1) - ln(x + 1)

. . . . . .y' . . . . . . . .1 . . . . . . . . . . . . . 2x . . . . 1
Then: . -- . = . (1/2)·-- + 5x^4 + (1/3)·-------- - ------- . . . . and so on . . .
. . . . . .y . . . . . . . . x . . . . . . . . . . . .x² + 1---x + 1

7. Thanks man, I really appreciate you taking the time to help me out.