Use the definition of the derivative to show that if f(x)= square root of x, then f(x)= 1/2(square-root x)
makes no sense
how can √x = (1/2)√x
only if x = 0
What does this have to do with a derivative?
its supposed to be f(x)= √x, then f '(x) = 1/2√x
if f(x) = √x = x^(1/2) , then
f ' (x) = (1/2)x^(-1/2)
= (1/2)(1/√x) , not what you typed.
To show that the derivative of f(x) = √x is f'(x) = 1/2√x using the definition of the derivative, we need to find the limit of the difference quotient as it approaches zero.
The definition of the derivative of a function f(x) at a point x is given by:
f'(x) = lim(h→0) [f(x + h) - f(x)] / h
Let's substitute the function f(x) = √x into this definition:
f'(x) = lim(h→0) [(√(x + h) - √x) / h]
To simplify this expression, we can use a common technique called "rationalizing the denominator." We multiply the numerator and denominator of the expression by the conjugate of the numerator, which is (√(x + h) + √x).
f'(x) = lim(h→0) [((√(x + h) - √x) / h) * (√(x + h) + √x) / (√(x + h) + √x)]
Expanding the numerator using the difference of squares and simplifying:
f'(x) = lim(h→0) [(x + h - x) / (h * (√(x + h) + √x))]
Canceling out the x terms in the numerator:
f'(x) = lim(h→0) [h / (h * (√(x + h) + √x))]
The h terms in the numerator and denominator cancel out:
f'(x) = lim(h→0) [1 / (√(x + h) + √x)]
Taking the limit as h approaches 0:
f'(x) = 1 / (√x + √x)
Combining the two square roots in the denominator:
f'(x) = 1 / (2√x)
Finally, simplifying the expression:
f'(x) = 1/2√x
So we have shown that the derivative of f(x) = √x is f'(x) = 1/2√x using the definition of the derivative.