DerivativeConsider an example of a square root function and find it's derivative using definition of derivative

Expert Answers
justaguide eNotes educator| Certified Educator

Let f(x) = sqrt x

f'(x) = lim h-->0 [ (sqrt (h + x) - sqrt x)/h]

=> lim h-->0 [ (sqrt (h + x) - sqrt x)/(h + x - x)]

=> lim h-->0 [ (sqrt (h + x) - sqrt x)/(sqrt (h + x))^2 - (sqrt x)^2)]

=> lim h-->0 [ (sqrt (h + x) - sqrt x)/(sqrt (h + x)-sqrt x)*(sqrt (h + x)+sqrt x)]

=> lim h-->0 [1/(sqrt (h + x)+sqrt x)]

substitute h = 0

=> 1/(sqrt x + sqrt x)

=> 1/2*sqrt x

This gives f'(x) = 1/2*sqrt x

giorgiana1976 | Student

Let's recall how to determine a derivatiove of a function using the first principle.

We'll express the first principle of finding the derivative of a given function:

lim [f(x+h) - f(x)]/h, for h->0

We'll consider the square root function f(x) = sqrt(3x+7)

We'll apply the principle to the given polynomial:

lim {sqrt [3(x+h)+7] - sqrt(3x+7)}/h

The next step is to remove the brackets under the square root:

lim [sqrt (3x+3h+7) - sqrt(3x+7)]/h

We'll remove multiply both, numerator and denominator, by the conjugate of numerator:

lim [sqrt (3x+3h+7) - sqrt(3x+7)][sqrt (3x+3h+7)+sqrt(3x+7)]/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll substitute the numerator by the difference of squares:

lim [(3x+3h+7) - (3x+7)]/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll eliminate like terms form numerator:

lim 3h/h*[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll simplify and we'll get:

lim 3/[sqrt (3x+3h+7)+sqrt(3x+7)]

We'll substitute h by 0:

lim 3/[sqrt (3x+3h+7)+sqrt(3x+7)] = 3/[sqrt(3x+7)+sqrt(3x+7)]

We'll combine like terms from denominator: