Question

Help with a few Calculus problems. WILL FAN AND MEDAL


I need to find the derivitave of a couple problems and simple can't figure them out

F (x)= 4x^4-(1/×^2)

F (x)=(1/4-x^2)

F (x)=(2x^3-3x^2+4x-5)/(x^2)

U (x)=1/(x+2) ^2

Answer 1

for the first one, it might help to think of 1/x^2 as x^(-2)

Answer 2

You'll use the power rule

if y = x^n then dy/dx = n*x^(n-1)

Answer 3

Would the first one be 16X^3+2x?

Answer 4

the derivative of x^(-2) is -2x^(-3)

notice how I pulled down the exponent and then subtracted 1 from -2 to get -3

Answer 5

-2x^(-3) = -2/(x^3)

Answer 6

So 16x^3+2x^3

Answer 7

more like \[\Large 16x^3 - \frac{2}{x^3}\]

Answer 8

Ah ok now I see

Answer 9

the second one isn't really clear, can you draw it out?

Answer 10

No I'm on my tablet but it's 1 divided by 4-x^2

Is that any better? I'm sorry!

Answer 11

ok so you'll need to use the quotient rule for this one

Answer 12

That's (low)(d'high)-(high)(d'low) right?

Answer 13

yeah I'm used to this formula


if
\[\Large h(x) = \frac{f(x)}{g(x)}\]
then
\[\Large h \ '(x) = \frac{f \ '(x)*g(x) - f(x)g \ '(x)}{[g(x)]^2}\]

Answer 14

Yes that's it. I can't figure it all out though it's confusing me

Answer 15

in this case

f(x) = 1
g(x) = 4-x^2

what are `f ' (x)` and `g ' (x)` equal to?

Answer 16

F (x)= 0
G (x)= 2x

Answer 17

That's the part that throws me

Answer 18

g ' (x) = -2x

Answer 19

Oh right

Answer 20

\[\Large h \ '(x) = \frac{f \ '(x)*g(x) - f(x)g \ '(x)}{[g(x)]^2}\]
\[\Large h \ '(x) = \frac{0*(4-x^2) - 1*(-2x)}{(4-x^2)^2}\]
now simplify

Answer 21

2x/(4-x^2)^2

Answer 22

yes

Answer 23

you'll use the quotient rule for `(2x^3-3x^2+4x-5)/(x^2)`

Answer 24

So

(X^2)(6x^2-6x+4)-(2x^3-3x^2+4x-5)(2x)

Answer 25

that's all over [g(x)]^2 = [x^2]^2 = x^4

Answer 26

Then I foil and combine what I can?

Answer 27

yes simplify as much as possible

Answer 28

(6x^4-6x^2+4x^2-4x^4-6x^3+8x^2-10x)

Answer 29

10x^4-2x^2+6x^3-10x

Answer 30

I'm not getting the same

Answer 31

What am I doing wrong?

Answer 32

(x^2)*(6x^2-6x+4)-(2x^3-3x^2+4x-5)*(2x)

(x^2)*(6x^2)+(x^2)*(-6x)+(x^2)*(4)-2x*2x^3-2x*(-3x^2)-2x*4x-2x*(-5)

6x^4-6x^3+4x^2-4x^4+6x^3-8x^2+10x

6x^4-4x^4-6x^3+6x^3+4x^2-8x^2+10x

2x^4-4x^2+10x

Answer 33

so you'll have 2x^4-4x^2+10x all over x^4

Answer 34

then you can factor out x from the numerator and have it cancel with one factor of x from the denominator

Answer 35

2x^3-4x+10

Answer 36

Over x^3

Answer 37

correct

Answer 38

It's always the factoring and foiling

Answer 39

So we use the quotient rule again?

(x+2)^2 (0)-(1)(2x)

Answer 40

Is it just 2x/(x+4)^2

Answer 41

Could you possibly post just one problem at a time? I do not readily recognize what your "Is it just 2x/(x+4)^2" refers to. A new problem? what are the instructions for it?

Answer 42

Just find the derivitave. I was referring to the last problem on the last after it's been simplfied