1. ## derivative ques

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

$
f (x) = x,
g (x) = \sqrt{x^2 + 1}
$

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

using the product rule:

$

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

$

this correcT? thank you

2. Originally Posted by jvignacio
$

y'(x) = \sqrt{x^2 + 1} + x\left(\frac{x}{\sqrt{x^2 + 1}}\right)

$

this correcT? thank you
Yes it is.

Laurent.

3. Originally Posted by jvignacio
$
y = x\sqrt{x^2+1}
$

$
f (x) = x,
g (x) = \sqrt{x^2 + 1}
$

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

using the product rule:

$

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

$

this correcT? thank you
Yes, it is correct, but it can be further solved. See here,

$= (\sqrt{x^2 + 1}) + \frac{x^2}{\sqrt{x^2 + 1}}$

$= \frac{x^2 + 1 + x^2}{\sqrt{x^2 + 1}}$

$= \frac{2x^2 + 1}{\sqrt{x^2 + 1}}$

4. Originally Posted by Shyam
Yes, it is correct, but it can be further solved. See here,

$= (\sqrt{x^2 + 1}) + \frac{x^2}{\sqrt{x^2 + 1}}$

$= \frac{x^2 + 1 + x^2}{\sqrt{x^2 + 1}}$

$= \frac{2x^2 + 1}{\sqrt{x^2 + 1}}$
thanks u!!!

5. You may not have learned this method yet but I think it is easier than remembering the product and quotient rules sometimes, logarithmic differentiation. If you would like me to show you more examples I be happy too.

$ln y = ln (x\sqrt{x^2+1})$

$ln y = lnx + ln\sqrt{x^2+1}$

$ln y = lnx +\frac{lnx^2+1}{2}$

now take the derivative

$\frac{y'}{y} = \frac{1}{x} + \bigg(\frac{1}{2}\bigg)\bigg
(\frac{2x}{x^2+1}\bigg)$

$\frac{y'}{y} = \frac{x^2+1+x^2}{x^3+x}$ = $\frac{2x^2 + 1}{x^3 + x}$

$y' = \bigg(y\bigg)\bigg(\frac{2x^2 + 1}{x^3 + x}\bigg)$

$y' = \bigg(\frac{x\sqrt{x^2+1}}{1}\bigg)\bigg(\frac{2x^ 2 + 1}{x^3 + x}\bigg)$ = $\frac{(\sqrt{x^2+1})(2x^2+1)}{x^2+1}$ = $\frac{2x^2+1}{\sqrt{x^2+1}}$

So

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