# Math Help - If y= log {x + (1 = x^2)^1/2} what is dy/dx?

1. ## If y= log {x + (1 + x^2)^1/2} what is dy/dx?

I hope someone can help me with this problem:

If y= log {x + (1 + x^2)^1/2} what is dy/dx?

apprantly the answer is dy/dx = (1 + x^2)^-1/2 , but I cannot see how to do it. Is it a function of a function? My lecturer showed us this on the board - but I still can't understand - so I would appreciate it if someone could help - thank you.

2. Originally Posted by goldilocks
I hope someone can help me with this problem:

If y= log {x + (1 + x^2)^1/2} what is dy/dx?

apprantly the answer is dy/dx = (1 + x^2)^-1/2 , but I cannot see how to do it. Is it a function of a function? My lecturer showed us this on the board - but I still can't understand - so I would appreciate it if someone could help - thank you.
Hello,

from the answer you gave I assume that you mean:

$f(x) = \ln(x+\sqrt{1+x^2})$ . Calculate the derivative using chain rule:

$f'(x) = \frac{1}{x+\sqrt{1+x^2}} \cdot \left({1+\frac12 \cdot (1+x^2)^{-\frac12} \cdot 2x}\right)$ = $\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$

Now rationalize(?) the denominator by multiplying this fraction by $\frac{x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}}$. You'll get:

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

3. Hello, goldilocks!

earboth did a lovely job . . . I used a slightly different simplication.

Differentiate: . $y \:= \:\ln\left[x + (1 + x^2)^{\frac{1}{2}}\right]$

Chain Rule: . $\frac{dy}{dx}\;=\;\frac{1}{x + (1+x^2)^{\frac{1}{2}}}\cdot\left[1 + \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot2x\right] \;=\;\frac{1 + \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$

Multiply top and bottom by $\sqrt{1+x^2}$

. . $\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\cdot\frac{1 + \frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} \;=\;\frac{\overbrace{\sqrt{1+x^2} + x}^{\text{cancel}}}{\sqrt{1+x^2}\underbrace{\left( x+\sqrt{1+x^2}\right)}_{\text{cancel}}} \;=\;\frac{1}{\sqrt{1+x^2}}$

4. Great - thank you both, that made it a lot clearer to see it like that. Thank you very much :-)

5. Originally Posted by goldilocks
If y= log {x + (1 + x^2)^1/2} what is dy/dx?
Have you ever hear about hyperbolic functions?, 'cause this is the $\mathrm{arcsinh}\,x$ function.

So $y=\mathrm{arcsinh}\,x\implies\sinh y=x$

Now let's differentiate

$y'\cosh y=1\implies y'=\frac1{\cosh y}$

Since $\cosh^2y-\sinh^2y=1$, yields

$y'=\frac1{\sqrt{1+x^2}}$