I understand finding the derivative and there is a simpler way of doing so, however, can someone show you derive at these answers by the DEF'N of Derivatives? Much appreciated!
Question 1: Find, from the definition, the derivative, f'(x), for f(x)=5x^2-3x+1 .
Answer: f ’(x) = 10x-3 .
Question 3: Find, from the definition, the derivative, f'(x), for f(x)=√(9x+4) .
Answer: f ’(x) = 9/(2√(9x+4)) .
f(x+h) = 5(x+h)^2 - 3(x+h) + 1
= 5x^2+10xh+5h^2 -3x-3h + 1
f(x) = 5x^2 -3x+1
subtract
f(x+h) - f(x) = 10 xh + 5 h^2 -3 h
divide by h to get slope
slope = 10 x +5 h - 3
let h go to zero to get slope at x
dy/dx = 10 x -3
now try the other one
f(x+h) = (9x+9h+4)^.5
[f(x+h)]^2 = 9 x + 9 h + 4
[f(x)]^2 = 9x+4
subtract and get 9h, remember that
a^2-b^2 = (a+b)*(a-b)
so
[f(x+h)]^2 - [f(x)]^2 =
{[f(x+h)]+[f(x)]}*{[f(x+h)]-[f(x)]}
SO
{[f(x+h)]-[f(x)]}={[f(x+h)]^2- [f(x)]^2} /{[f(x+h)]+[f(x)]}
ah ha, I want what I see on the left
now from way up there I know
{[f(x+h)]^2- [f(x)]^2} = 9h
so
{[f(x+h)]-[f(x)]}=9h /{[f(x+h)]+[f(x)]}
divide by h to get slope
9 /{[f(x+h)]+[f(x)]}
let h --->0
9/{2 f(x)}
= 9/{2sqrt(9x+4)}
so I agree with you
I did the second one because it is much harder than the first one and if you had trouble with the first you were in much greater trouble with the second.
Certainly! In order to find the derivatives of these functions using the definition of derivatives, we need to use the limit definition of the derivative.
The definition states that the derivative of a function f at a point x is given by the limit as h approaches 0 of [f(x+h) - f(x)] / h.
Let's start with Question 1, where we have f(x) = 5x^2 - 3x + 1. To find f'(x) using the definition, we apply the limit:
f'(x) = lim (h -> 0) [f(x + h) - f(x)] / h
Substituting f(x) in the equation, we get:
f'(x) = lim (h -> 0) [(5(x+h)^2 - 3(x+h) + 1) - (5x^2 - 3x + 1)] / h
Expanding and simplifying:
f'(x) = lim (h -> 0) [(5x^2 + 10xh + 5h^2 - 3x - 3h + 1) - (5x^2 - 3x + 1)] / h
Canceling out like terms:
f'(x) = lim (h -> 0) [10xh + 5h^2 - 3h] / h
Factoring out h:
f'(x) = lim (h -> 0) h(10x + 5h - 3) / h
Now, canceling out h:
f'(x) = lim (h -> 0) 10x + 5h - 3
Finally, as h approaches 0, the term 5h becomes negligible, giving us:
f'(x) = 10x - 3
And that's the derivative of f(x) = 5x^2 - 3x + 1 using the definition of derivatives!
Now let's move on to Question 3, where we have f(x) = √(9x + 4). Again, we'll apply the definition of derivatives.
Using the same steps as before, we find:
f'(x) = lim (h -> 0) [√(9(x + h) + 4) - √(9x + 4)] / h
Simplifying, we get:
f'(x) = lim (h -> 0) [√(9x + 9h + 4) - √(9x + 4)] / h
Now, in order to eliminate the square root, we can multiply the numerator and denominator by the conjugate of the numerator:
f'(x) = lim (h -> 0) [(√(9x + 9h + 4) - √(9x + 4)) * (√(9x + 9h + 4) + √(9x + 4))] / h * (√(9x + 9h + 4) + √(9x + 4))
Expanding and simplifying:
f'(x) = lim (h -> 0) [(9x + 9h + 4) - (9x + 4)] / h * (√(9x + 9h + 4) + √(9x + 4))
Canceling out like terms:
f'(x) = lim (h -> 0) [9h] / h * (√(9x + 9h + 4) + √(9x + 4))
Canceling out h:
f'(x) = lim (h -> 0) 9 * (√(9x + 9h + 4) + √(9x + 4))
Finally, as h approaches 0, the term √(9x + 9h + 4) + √(9x + 4) no longer has a dependence on h, so we obtain:
f'(x) = 9/(2√(9x + 4))
And that's the derivative of f(x) = √(9x + 4) using the definition of derivatives!
I hope that helps clarify the process of finding derivatives using the definition.