# Math Help - A few basic calculus derivative problems

1. ## A few basic calculus derivative problems

1. If f(x)=3(root x)*(x^3-7(root x) +7)

find f'(x) and f'(4)

2. If f(x)=9x+2 / x^2+2x-3, find the derivative f'(x).

3. G(x)=(3x^2-2)(x+2)^2 / (10x^2+8x+2)

Find G'(2)

4. f(x)=x^8 h(x)

h(-1)=3
h'(-1)=6

Calculate f(-1)

I attempted all of these, thought I was doing everything correct but apparently I was not and now im stuck. Any help is greatly appreciated

2. Edit, I just found number 2 so scratch that one

3. Originally Posted by sitdownson
1. If f(x)=3(root x)*(x^3-7(root x) +7)

find f'(x) and f'(4)
1. $3\sqrt{x} \cdot (x^3 - 7\sqrt{x} +7)$

Distribute the $3\sqrt{x}$ to get $f(x) = 3x^{3.5} - 21x + 21x^{0.5}$

You can differentiate that one in the standard way - please post any workings you've done if you have any problems

4. ok great, got numbers 1 and 2 now, still trying to figure out what i did wrong on the rest, but thanks for the help so far!

6. Ok sorry about that first time on these forums....

3. f'(x) 2x +4 (6x) / 20x + 8

f'(2) (4+4)(12) / 40 +8
96/48
2
Got this wrong

4. I thought that h(x) was 3x^2 but I realized I was wrong and now I have no idea what to do

7. #3 ...

$\displaystyle G(x) = \frac{1}{2} \cdot \frac{(3x^2-2)(x+2)^2}{5x^2+4x+1}$

$\displaystyle G'(x) = \frac{1}{2} \cdot \frac{(5x^2+4x+1)[(3x^2-2) \cdot 2(x+2) + (x+2)^2 \cdot 6x] - (3x^2-2)(x+2)^2(10x+4)}{(5x^2+4x+1)^2}$

now evauate $G'(2)$

#4 ...

$f(x) = x^8 \cdot h(x)$

$f'(x) = x^8 \cdot h'(x) + h(x) \cdot 8x^7$

now evaluate $f'(-1)$

8. Originally Posted by sitdownson
3. G(x)=(3x^2-2)(x+2)^2 / (10x^2+8x+2)

Find G'(2)
It looks like you have found G(2) rather than G'(2)
$\dfrac{(3x^2-2)(x+2)^2}{10x^2+8x+2} = \dfrac{1}{2} \cdot \dfrac{(3x^2-2)(x+2)^2}{5x^2+4x+1}$

Here you need to use the product rule and quotient rule. First the quotient rule

$u = (3x^2-2)(x+2)^2$ and $v = 10x^2+8x+2$

To find u' we need to use the product rule - for clarity's sake I'm going to use p and q

$p = 3x^2-2$ and $q = (x+2)^2$

$u' = p'q + q'p = 6x(x+2)^2 + 2(x+2)(3x^2-2)$

Now we have found u' we can use it in the quotient rule for the whole expression:

$u' = 6x(x+2)^2 + 2(x+2)(3x^2-2)$
$v' = 20x +8$

$g'(x) = \dfrac{vu'-uv'}{v^2} = \dfrac{[(10x^2+8x+2)(6x(x+2)^2 + 2(x+2)(3x^2-2)] - [(3x^2-2)(x+2)^2(20x+8)]}{(10x^2+8x+2)^2}$

$g'(2) = \dfrac{(40+16+2)(12(4)^2 + 2(6)(10))] - [(10)(16)(48)]}{(40+16+2)^2} = \dfrac{58 \cdot 312 - 7680}{3364} = \dfrac{2604}{841} \approx 3.096$