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)
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???
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 $\displaystyle \frac{d}{\,dx}\left[x^{\frac{3}{2}}\right]\left(3\ln(x)-2\right)+x^{\frac{3}{2}}\cdot\frac{d}{\,dx}\left[3\ln(x)-2\right]=\dots$
$\displaystyle \frac{dy}{dx}=\sqrt{x}(3ln(x)-2)+\frac{x(3ln(x)-2)}{2\sqrt{x}}+\frac{3x\sqrt{x}}{x}$
Remember $\displaystyle \frac{d(uv)}{dx}=u\frac{d(v)}{dx}+ v\frac{d(u)}{dx}$
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
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