1. ## derivative

Find the derivatives of the follwoing functions :

a) y= In(x + squareroot of (x^2 + 1));

b) y = x * Squareroot of x * (3In x -2)

Originally Posted by Faz
Find the derivatives of the follwoing functions :

a) y= In(x + squareroot of (x^2 + 1));
a) $\displaystyle y = \ln (x+ \sqrt{x^2+1})$

$\displaystyle y'=\frac{1}{x+ \sqrt{x^2+1}}.\left(1+\frac{1}{2\sqrt{x^2+1}}.2x\r ight)$

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

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

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

Did you get it now???

3. Originally Posted by Faz
Find the derivatives of the follwoing functions :

b) y = x * Squareroot of x * (3In x -2)
Is your question like this ??

$\displaystyle y=x\sqrt {x} (3\ln x -2)$

4. ## yes

yes very correct

5. Originally Posted by Faz
find derivatives of function

Note that $\displaystyle x\sqrt{x}=x^{\frac{3}{2}}$

Thus, apply the product rule to $\displaystyle x^{\frac{3}{2}}\left(3\ln(x)-2\right)$. You should get $\displaystyle \frac{d}{\,dx}\left[x^{\frac{3}{2}}\right]\left(3\ln(x)-2\right)+x^{\frac{3}{2}}\cdot\frac{d}{\,dx}\left[3\ln(x)-2\right]=\dots$

6. $\displaystyle \frac{dy}{dx}=\sqrt{x}(3ln(x)-2)+\frac{x(3ln(x)-2)}{2\sqrt{x}}+\frac{3x\sqrt{x}}{x}$
Remember $\displaystyle \frac{d(uv)}{dx}=u\frac{d(v)}{dx}+ v\frac{d(u)}{dx}$

7. Originally Posted by Faz
how do i solve this ?

y=x*squareroot of X(3lnx-2)
Hi man,

why is there a big X?

$\displaystyle y = x * \sqrt{x*(3ln(x)-2)}$

$\displaystyle u:= x$

$\displaystyle u '=1$

$\displaystyle v:= \sqrt{x*(3ln(x)-2)} = (x*(3ln(x)-2))^0.5$

$\displaystyle a:= x$
$\displaystyle b:= 3ln(x)-2$
$\displaystyle a' = 1$
$\displaystyle b' = 3/x$

=> $\displaystyle v' = 0.5*[x*3/x + 1*(3ln(x)-2)]^{-0.5}$

$\displaystyle = 0.5 * [3 + 3ln(x) -2]^{-0.5} = 0.5* [3*ln(x)+1]^{-0.5}$

=> y' = u*v' + v' * u (Product rule)

Can you solve the rest? (I already told you u, u', v, v'). I guess you can

Regards
Rapha

8. Originally Posted by Faz
y=x*squareroot of x * (3lnx-2)
The question is:

Find the derivative of:

$\displaystyle y=x \sqrt{x}(3 \ln(x)-2)$

This involves a certain amount of guess work about where to put the brackets.

Rewrite this as:

$\displaystyle y=x^{3/2}(3 \ln(x)-2)$

Then we need the product rule:

$\displaystyle \frac{dy}{dx}=\left[ \frac{d}{dx}x^{3/2}\right](3 \ln(x)-2)+x^{3/2}\left[\frac{d}{dx}(3 \ln(x)-2)\right]$

CB