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.