use the limit definition of the derivative to find f'
a. let f(x)= 1/(x^2)
b. let f(x)= x^4
c. let f(x)= 9(3√(x^2)
(a) is clearly explained at
http://tinypic.com/view.php?pic=15qynhv&s=4#.VO_LNywYFvA
You can probably find the others as well. Usually the Binomial Theorem is involved.
For instance,
(b) f(x+h) - f(x)
= (x+h)^4 - x^4
= x^4+4x^3h+6x^2h^2+4xh^3+h^4 - x^4
= 4x^3h+6x^2h^2+4xh^3+h^4
Now divide that by h, take the limit, and all the h stuff disappears, leaving you with 4x^3
a. To find the derivative of f(x) = 1/(x^2) using the limit definition of the derivative, we need to first write out the definition. The limit definition of the derivative states:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Now, substitute f(x) = 1/(x^2):
f'(x) = lim(h->0) [1/((x + h)^2) - 1/(x^2)] / h
Next, simplify the expression inside the limit:
f'(x) = lim(h->0) [(x^2 - (x + h)^2) / (x^2((x + h)^2))] / h
Expand (x + h)^2:
f'(x) = lim(h->0) [(x^2 - (x^2 + 2hx + h^2)) / (x^2(x^2 + 2hx + h^2))] / h
Simplify the numerator:
f'(x) = lim(h->0) [-2hx - h^2] / (x^2(x^2 + 2hx + h^2)) / h
Cancel out the common factor of h:
f'(x) = lim(h->0) [-2x - h] / (x^2(x^2 + 2hx + h^2))
Take the limit as h approaches 0:
f'(x) = -2x / (x^2(x^2))
Simplify further:
f'(x) = -2 / (x^3)
Therefore, the derivative of f(x) = 1/(x^2) is f'(x) = -2 / (x^3).
b. To find the derivative of f(x) = x^4 using the limit definition of the derivative, we apply the same steps as above:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Substitute f(x) = x^4:
f'(x) = lim(h->0) [(x + h)^4 - x^4] / h
Expand (x + h)^4:
f'(x) = lim(h->0) [(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4] / h
Simplify the numerator:
f'(x) = lim(h->0) [4x^3h + 6x^2h^2 + 4xh^3 + h^4] / h
Cancel out the common factor of h:
f'(x) = lim(h->0) [4x^3 + 6x^2h + 4xh^2 + h^3]
Take the limit as h approaches 0:
f'(x) = 4x^3 + 0 + 0 + 0
Simplify further:
f'(x) = 4x^3
Therefore, the derivative of f(x) = x^4 is f'(x) = 4x^3.
c. To find the derivative of f(x) = 9(3√(x^2)), we can simplify the expression before applying the limit definition of the derivative:
f(x) = 9(3√(x^2))
= 27√(x^2)
= 27|x|
Now, we can find the derivative of f(x) = 27|x|:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Substitute f(x) = 27|x|:
f'(x) = lim(h->0) [27|x + h| - 27|x|] / h
Split the function into two cases, one for x≥0 and the other for x<0:
For x≥0:
f'(x) = lim(h->0) [27(x + h) - 27x] / h
= lim(h->0) [27x + 27h - 27x] / h
= lim(h->0) 27h / h
= lim(h->0) 27
= 27
For x<0:
f'(x) = lim(h->0) [27(-x - h) - 27(-x)] / h
= lim(h->0) [-27x - 27h + 27x] / h
= lim(h->0) -27h / h
= lim(h->0) -27
= -27
Therefore, the derivative of f(x) = 9(3√(x^2)) is f'(x) = 27 for x≥0 and f'(x) = -27 for x<0.