Girlfriend needs help with these ASAP, and I haven't learned that type of math. Looking for skype calls maybe, or any way to help. Can pay if you wants it

If you post the problem I may be able to help, don't want to jump on a skype call though.

I have no idea about any of this, but here is her page, I think she said 2(b).

I don't see a 2(b), but it looks like she did 2.1 correctly. The rest of these are almost all done with the same technique, you just need to rewrite the function to an easier format.

2.2 is done the same way: y' = 15x^4 + 21x^2 - 10
2.3 is done the same way: y' = 16x^3 - 9x^2 - x

Easiest way to do 2.4 is to rewrite the equation using negative exponents, and then proceed the same as in 2.1...multiplying by the exponent and subtracting one from it.

f(k) = -4k^-1 + 6k^-3 - 10k^-5 + 12^.5
f'(k) = 4k^-2 - 18k^-4 + 50k^-6

2.5 is done the same as 2.1: f'(t) = (2.5 * 5)t^1.5 - (8 * 0.5)t^-0.5 Not sure if that exponent is 0.5 or 6.5

2.6: Once again, rewrite the function
y = -15x^.5 - 12x^(4/8)
y' = -7.5x^-.5 - 6x^-.5

2.8: This one is going to be using the chain rule, rewrite it with a negative exponent:
y = 10 * (3 - 2x)^-4
y' = (-4) * 10 * (3 - 2x)^-5 * (-2)
y' = 80 * (3 - 2x) ^ -5
y' = 80 / (3 - 2x)^5


2.9: Rewrite it to have an exponent instead of square root, and then do chain rule again.
y = (x^3 - 7x^2 + 5 x + 2)^.5
y' = 0.5 * (x^3 - 7x^2 + 5x + 2) ^ (-0.5) * (3x^2 - 14x + 5)
y' = (0.5 * (3x^2 - 14x + 5)) / (x^3 - 7x^2 + 5x + 2) ^.5

y' = (1.5x^2 - 7x + 2.5) / (x^3 - 7x^2 + 5x + 2)^.5



Whoops, skipped 2.7

Chain rule once again, or if you wanted, you could just do FOIL to have an expanded function and just do it the same as 2.1, but doing chain rule style:
f'(g) = 2 * 4 * (3g - 5) * 3
f'(g) = 24 * (3g - 5)


Incase you meant 1(b):
f(x) = 4x^2 - 6x
f(x+h) = 4(x+h)^2 6(x+h) = 4x^2 + 4h^2 + 8hx - 6x - 6h
f(x + h) - f(x) = 4h^2 + 8hx - 6h

( f(x + h) - f(x) ) / h = 4h + 8x - 6
lim (h -> 0 ) of the above = 8x - 6

f'(-2) = -16 - 6 = -22
f'(0) = -6
f'(3) = 24 - 6 = 18

This post was edited by RzChaos on Nov 13 2015 06:26pm

for 2.3 how did you get X from X^2/2

The exponent was 2, so subtract 1 and now the exponent is 1
Multiply x/2 by 2

Awesome That makes sense. Feel so stupid after realizing. And for 2.4 could you explain how to change it into the negative exponents. I haven't done this math since high school

If you have a variable in the denominator, you can move into the numerator by changing the sign of the exponent.

So 1 / X = X^ -1

1 / x^-2 = X^2

1 / (X + 3)^2 = (X + 3)^-2

You can do the same to move something to the denominator.

X^-2 = 1 / X^2

This post was edited by RzChaos on Nov 13 2015 07:01pm

Okay awesome. I understand that. Now for 2.4 still what about the square root? wouldn't that be 3.4

The derivative of a "plain" number is 0, so it disappears.

The numbers only stick around if they are connected to a variable in some manner.

Just like how the 8 disappeared in 2.1

This post was edited by RzChaos on Nov 13 2015 07:23pm