Substitution and chain rule

[Bobby Jones]

Using the substitution and chain rule, is this correct:


Ln(x + (x^2 + 1)^1/2) = (x+1)/x^2


Am I right or wrong because the squareroot over the x^2 + 1, confused me a little.


Cheers guys

 

[soroban]

Hello, Bobby Jones!

Using the Substitution and Chain Rule, is this correct?

\(\displaystyle f(x) \:=\:\ln\left[x + \left(x^2 + 1\right)^{\frac{1}{2}}\right]\)

\(\displaystyle f'(x) \:=\:\frac{x+1}{x^2}\) . Sorry, it's wrong.

I just used the Chain Rule.

\(\displaystyle f'(x) \;=\;\frac{1}{x + \sqrt{x^2+1}}\cdot\left[1 + \tfrac{1}{2}(x^2+1)^{-\frac{1}{2}}\cdot2x\right]\text{ }\)

. . . . .\(\displaystyle =\;\frac{1}{x + \sqrt{x^2+1}} \cdot\left(1 + \frac{x}{\sqrt{x^2+1}}\right)\)

. . . . .\(\displaystyle =\;\frac{1}{x + \sqrt{x^2+1}} \cdot\left(\frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right)\)

. . . . .\(\displaystyle =\;\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}\cdot(x+\sqrt{x^2+1})}\)

. . . . .\(\displaystyle =\;\;\frac{1}{\sqrt{x^2+1}}\)

 

[Bobby Jones]

hmm, I'm looking through your work, and I can see where I went wrong. I take it you did the chain rule twice in effect.
When I look at the way you have done it, it looks so simple and makes sense.


Good job man

Cheers