Question

Derivate:
f(x)=ln(sqrt(x) + sqrt(x+1))
f'(1) = ?

I don't really understand wolframs steps.
So could anyone help me :(.

Answer 1

...

Answer 2

chain rule... first let x = (sqrt x + sqrt x +1

so 1/x(dx)

dx is the derivative of x which is sqrt x + sqrt x+1

take the derivative of that...
1/2sqrtx + 1/2sqrt(x+1)

so your answer is

(1/2sqrtx + 1/2sqrtx+1)/x

simplify the numerator...
1/2sqrtx + 1/2sqrtx+1

LCD = 2(sqrtx)(sqrtx+1)

so sqrt(x+1) + sqrtx/2(sqrtx)(sqrtx+1)/x

simplify...

sqrt(x+1) + sqrtx/2x(sqrtx)(sqrtx+1)

was that what wolframalpha gave you?

Answer 3

oh...forgot to sub...
sqrt2 + 1/2sqrt2

Answer 4

wolfram says:
1/(2 sqrt(x) sqrt(1+x))

Answer 5

oh i'm sorry!!! i missubstituted...i made the denominator x when it was supposed to be sqrtx + sqrtx+1...lemme redo that...

Answer 6

\[\large \log(\sqrt{x} +\sqrt{x+1}) \]
Use:
\[\frac{d}{dx} \log(..) = \left(\frac{1}{(..)}\right) * (..)'\]
When you find the derivative of: \[\sqrt{x} +\sqrt{x+1} \] =>\[\frac{1}{2}\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x}}\right) \]

\[\frac{1}{\sqrt{x}+\sqrt{x+1}} *\frac{1}{2}\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x}}\right) \]

Maybe..It looks like it might match Wolfram Alpha's Answer.

Answer 7

@mimi, it's not log, it's natural logarithm. .

Answer 8

loolol, typo.

Answer 9

but its still the same answer i suppose.