1. Differentiation problems.

Differentiate the following:
1. $f(x) =\sqrt{x}(2x+3)^{2}$
2. $f(x) = \frac{sinx}{x^3-2x}$

My attempt for 1:
First I rewrite to get rid of the radical:
$f(x) = x^{\frac{1}{2}}(2x+3)^{2}$
Then I start differentiating using product rule:
$f'(x) = x^{\frac{1}{2}}'(2x+3)^{2} +x^{\frac{1}{2}}(2x+3)^{2}'$
Then chain rule:
$f'(x) = (\frac{1}{2})(x^{\frac{1}{2}-1)}(1)(2x+3)^{2} +(x^{\frac{1}{2}})(2)(2x+3)^{(2-1)}(2)$
$f'(x) = (\frac{1}{2})(x^{\frac{-1}{2})}(1)(2x+3)^{2}+(x^{\frac{1}{2}})(4)(2x+3)$
After simplifying:
$f'(x) = \frac{(2x+3)^{2}}{\sqrt{2}\sqrt{x}} + \frac{2x+3}{2\sqrt{x}}$

My attempt for number 2:
First I rewrite the equation to get rid of the fraction:
$f(x) = (sinx)(x^3-2x)^{-1)$
Then I begin to differentiate using product rule:
$f'(x) = (sinx)'(x^3-2)^{-1} +(sinx)(x^3-2x)^{-1}'$
Chain rule:
$(cosx)(x^3-2x)^{-1} + (sinx)(-1)(x^3-2x)^{-1-1}(3x^2-2)$
After simplifying:
$f'(x) = \frac{cosx}{(x^3-2x)} + \frac{2(3x^2 sinx)}{(x^3-2x)^{2}}$

What have I done wrong in my calculations? To check that I'm correct, I plug in random and find the slopes of the tangent to those points and check if the functions have been derived correctly through a graphing program. Unfortunately, I am way off. Thanks for any help in advance.

2. Hello, Pupil!

You made silly mistakes in the last step . . .

Differentiate the following:

$1.\;f(x) \:=\:\sqrt{x}(2x+3)^{2}$

$2.\;f(x) \:=\:\dfrac{\sin x}{x^3-2x}$

My attempt for 1:

First I rewrite to get rid of the radical: . $f(x) \:=\: x^{\frac{1}{2}}(2x+3)^{2}$

Then I start differentiating using product rule:
$f'(x) \:=\: x^{\frac{1}{2}}'(2x+3)^2 +x^{\frac{1}{2}}(2x+3)^2'$

Then chain rule:
$f'(x) \:=\: (\frac{1}{2})(x^{-\frac{1}{2}})(2x+3)^2+(x^{\frac{1}{2}})(4)(2x+3)$

After simplifying:

$f'(x) = \dfrac{(2x+3)^{2}}{\sqrt{2}\sqrt{x}} + \dfrac{2x+3}{2\sqrt{x}}$
. . . . . . . . . . . . . . . . . . .
. . . . . . . just 2. . . . . . in numerator

My attempt for number 2:

First I rewrite the equation to get rid of the fraction:
$f(x) = (\sin x)(x^3-2x)^{\text{-}1}$

Then I begin to differentiate using product rule:
$f'(x) = (\sin x)'(x^3-2)^{\text{-}1} +(\sin x)(x^3-2x)^{\text{-}1}'$

Chain rule:
$(\cos x)(x^3-2x)^{\text{-}1} + (\sin x)(-1)(x^3-2x)^{\text{-}2}(3x^2-2)$

After simplifying:. . . . . . .?
. . . . . . . . . . . . . . . . . . .
$f'(x) \:=\: \dfrac{\cos x}{(x^3-2x)} + \dfrac{2(3x^2-2)\sin x}{(x^3-2x)^{2}}$
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . minus

3. Hi Soroban, thanks for the reply.

Okay, so for 1:
$f'(x) = \frac{(2x+3)^{2}}{2\sqrt{x}} + \sqrt{x}(4)(2x+3)$ is this correct?

And for 2:
$f'(x) = \frac{cosx}{(x^3-2x)} - \frac{(3x^2 - 2sinx)}{(x^3-2x)^{2}}$
?