Question

calculate the derivative using first principle definition of derivative:
a)f(x)=x^2-2x+1
b)f(x)=x^3-3x^2
c)f(x)=sq root of 2x
d)f(x)=4/x+1
*show steps

2.Explain in own words the process used to conclude that the derivative of a cubic function is a quadratic one

Answer 1

a) f(x)=x^2-2x+1
f(x) = (x-1)^2
f'(x) = lim h->0 { f(x+h) - f(x) } / { (x+h) - x } = lim h->0 { (x+h-1)^2 - (x-1)^2 } h
= lim h->0 { (x+h-1+x-1)(x+h-1-x+1) } / h = lim h->0 (2x-2+h)(h)/h
= lim h->0 (2x - 2 + h) = 2x - 2

Answer 2

how to get d and number 2?

Answer 3

d) Is it 4/(x+1) or 4/x + 1 ?
Assuming it is f(x) = 4/(x+1)
f'(x) = lim h->0 { 4/(x+h+1) - 4/(x+1) } / h =
lim h->0 { 4(x+1) - 4(x+h+1) } / h(x+1)(x+h+1) = lim h->0 { -4h } / { h(x+1)(x+h+1) }
lim h->0 ( -4 ) / (x+1)(x+h+1) = -4/(x+1)^2

2) The general cubic function is: f(x) = ax^3 + bx^2 + cx + d
The derivative f'(x) = 3ax^2 + 2bx + c which is a quadratic function.

(Not sure if they are asking to prove this using the first principle).

Answer 4

how to prove it if using first principle?

Answer 5

Start with the general cubic function: f(x) = ax^3 + bx^2 + cx + d
Find f(x+h) and subtract f(x) from it. Here you will notice the cubic term cancels out but the quadratic term does not cancel out and it remains when we find the limit:
lim h->0 { f(x+h) - f(x) } / h
If they just want you to explain in your own words you can stop it at this point or you can go ahead and find the limit and you would have proven using first principle that the derivative of a cubic function is quadratic.

Answer 6

how to do the f(x)=square root of 2x?

Answer 7

f(x)=(1/2)(2x)^1/2-1)
f(x)=2x^-1/2
f(x)=1/2x^1/2

is this right?

Answer 8

what is b?

Answer 9

You have to do each one of the problems using the first principle much like a) was done.

Answer 10

For c) First setup the first-principle definition of the derivative. Then multiply the top and bottom by the conjugate. (that is if the top is: (sqrt(a) - sqrt(b)) then multiply top and bottom by (sqrt(a) + sqrt(b)) and simplify.