Question

Consider the function
G(x) = 4x^2 - x^3
(a) Find G ' (a).

Answer 1

polpok i need u!!!!1

Answer 2

Well start by taking the derivative of G. Then plug in a for x.

Answer 3

can you show me step by step i know the derivate is \[\lim h \rightarrow 0 f(a+h)-f(a) \div h\]

Answer 4

That's the definition of the derivative.. Are you supposed to be using the definition? or just the normal power rules?

Answer 5

well we must remember the power rules(or we could apply the method of increments to each term). So for the first term 4x^2 we bring the 2 as a multiplicative of the coefficient 4 thus making the coefficient 8. Then we subtract 1 from the original exponent so 2-1 equals 1. The derivative of the first term is 8x.

Answer 6

The second term x^3, we bring the exponent down as a coefficient. This gives us 3x^3 but now we subtract 1 from 3 so we have 3x^2. Thus altogether we have g'(x)= 8x-3x^2

Answer 7

i still dont understand cuz i know you have to use the deruvative but not sure how to use it

Answer 8

The derivative is the G'(x). For example the derivative of 4x^2 is 8x. And the derivative of x^3 is 3x^2

Answer 9

Do you know what the derivative actually is?

Answer 10

no not really

Answer 11

Okay well, the derivative is the instantaneous rate of change of y with respects to x. Do you understand that?

Answer 12

yes..but the other problem similar shows me something complety different with the use of you f(a+h)-F(a)/h

Answer 13

so your saying i just compute it

Answer 14

ok i understand, the way that you wish to use i also valid, but in the end it will give you the same answer. Your method is what you would call an introductory method to finding the derivative. This is perfectly fine, because with out this method we would not understand later or more sophisticated methods of getting the derivatives. Let me see if i can remember how to you the method you want to use.

Answer 15

ok

Answer 16

\[G'(x) = \lim_{h \rightarrow 0} {G(x+h) - G(x) \over h}\]
\[=\lim_{h \rightarrow 0} {4(x+h)^2 - (x+h)^3 - 4x^2 - x^3 \over h}\]
\[= \lim_{h \rightarrow 0} {4(x+h)^2 - (x+h)^3 - 4x^2 - x^3 \over h}\]

Then multiply out the squared and cubed terms and simplify.

Answer 17

ok this one is really confuse...so the way you got 8x from4x^2...did yo double it?

Answer 18

Did you multiply out the exponents in the numerator?

Answer 19

we must expand all the terms first and then there ill be some cancellations that we can do

Answer 20

yes...but ti your situation where you simplyfy the equation 4x^2-x^3= 8x-3x^2..i like to know how you figure the 8x

Answer 21

you will see.. expand out those terms first.

Answer 22

we figure 8x out using the power rules, but this come later in calculus, once you have mastered this particular method of finding the derivative.

Answer 23

Dont you want to find out how to find the derivative using the method you wanted to use?

Answer 24

we must expand the terms

Answer 25

after expansion you will see where the 8x comes from

Answer 26

yes but how did you find it by the first step you conclude
Also how would i solve it for the point (3,9)

Answer 27

i used the power rule which says g(x)=x^n=g'(x)=nx^n-1

Answer 28

so (2)4x^2-1=so ....is that it

Answer 29

if you haven't learned the power rule you should stick to the definition of the derivative.

Answer 30

yes, but do you understand what that means

Answer 31

most college students look a the power rule and say wow that easy computation, but they dont understand that it is actually describing something. Do the expansion of the terms and see how and why the power rule works

Answer 32

ok..well thanks for the help..ill try to do my best