Thread: calc help on integrals/differentiation

can anyone just double check these for me please??

differentiation

ln (radical((x^2) +4)) = x / (x^2) + 4

ln (5-x)^6 = -6/(5^-x)

ln ( x * (e^x) ) = (e^-x)((e^x)x + e^x )) / x

ln (e^x) / radical(e^x) + 1 = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2

can anyone just double check these for me please??

differentiation

ln (radical((x^2) +4)) = x / (x^2) + 4

ln (5-x)^6 = -6/(5^-x)

ln ( x * (e^x) ) = (e^-x)((e^x)x + e^x )) / x

ln (e^x) / radical(e^x) + 1 = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2

Hello, darkblue!

If no one is responding, it's because your work is hard to read.

. . . . . . - - - . ______
1. f(x) .= .ln(√x² + 4)
. . . . . . - - - . ._____
. . f'(x) .= .x /√x² + 4 . Right!


2. f(x) .= .(5 - x)^6

. . f'(x) .= .-6/(5 - x) . Yes!


3. f(x) .= .ln(x·e^x)

. . f'(x) .= .(e^-x)(x·e^x + e^x ) / x . ??

Where did the e^{-x} come from?

We have: .f(x) .= .ln(x·e^x) .= .ln(x) + ln(e^x) .= .ln(x) + x

. . Therefore: .f'(x) .= .1/x + 1 .= .(1 + x)/x

4. f(x) = ln (e^x) / radical(e^x) + 1
. . f'(x) = 1 / (radical(e^x) + 1) - radical(e^x) log (e^x) / 2(radical(e^x) +1)^2 . ??

Did you know that: ln(e^x) = 1 ?

. . . . . . . . . . . . . . . . . . . . . . 1
If the function is: .f(x) .= . ------------ .= .(e^x + 1)^{-½}
. . . . . . . . . . . . . . . . . . . √e^x + 1

. . the derivative is: .f'(x) .= .(-½)(e^x)·(e^x + 1)^{-3/2}

Remember that



If you apply this, with the first one, you've got:



Do the same with the others.

P.S.: this forum hasn't LaTeX?

Originally Posted by Krizalid

P.S.: this forum hasn't LaTeX?

It's a work in progress...

-Dan

Great!

Thanks for the notice.