# Derivative of Sqrt(x) With Limits

TR Smith is a product designer and former teacher who uses math in her work every day.

The first course in calculus is all about derivatives, how to calculate them and how to use them to solve a variety of math problems that require more than simple algebra or geometry. There are basically two ways you learn to differentiate functions in Calc I, the formal and hard way with limits, and the easy way with memorized formulas. The shortcut derivative formulas you memorize actually come from the limit method of finding derivatives.

For instance, most calculus students know that the derivative of a power function f(x) = x^n is f'(x) = n*x^(n-1), and that the derivative of a function sum g(x) + h(x) is g'(x) + h'(x). These formulas were not plucked from midair, but derived from the limit definition of the derivatives.

The function f(x) = sqrt(x) can be written as a power function f(x) = x^(1/2), which means that its derivative is f'(x) = (1/2)*x^(-1/2) = 1/[2*sqrt(x)] using the power rule. But if you don't yet know the power rule shortcut, how do you find the derivative of sqrt(x) using limits? This tutorial will show you step by step how to differentiate sqrt(x) with limits.

## Applying the Limit Definition of Derivative

The formal definition of a derivative says that if f(x) is a function, then its derivative f'(x) is given by the limit of the fraction

lim(h→0) [f(x+h) - f(x)]/h

Plugging h = 0 directly into this expression gives you the indeterminate form 0/0, so you always need to apply some algebra to rework the expression into something that allows you to plug in h = 0. Replacing f(x) with sqrt(x) gives you

lim(h→0) [sqrt(x+h) - sqrt(x)]/h

Now we will see how to simplify this expression.

## Simplifying lim(h→0) [sqrt(x+h) - sqrt(x)]/h

Multiplying the numerator and denominator of the fraction by the same expression does not change the fraction's value, therefore we can choose something that will eliminate the difference of square roots in the numerator. Let's try multiplying the top and bottom by sqrt(x+h) + sqrt(x). This gives us

lim(h→0) [sqrt(x+h) - sqrt(x)]/h

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

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

What happened in the line immediately above was that the square roots disappeared from the numerator. This is called rationalizing the numerator. (If you make square roots disappear from the denominator with this trick, it's called rationalizing the denominator). The essence of the trick is that whenever you see sqrt(a) - sqrt(b) in a fraction, you should multiply by sqrt(a) + sqrt(b) to get a - b. And if you see sqrt(a) + sqrt(b) you should multiply by sqrt(a) - sqrt(b) to get a - b. Continuing with simplifying the limit expression you get

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

At this point we can cancel out the factors of h from the numerator and denominator.

= lim(h→0) 1 / [sqrt(x+h) + sqrt(x)]

And now we can safely substitute h = 0 into the expression without getting an indeterminate form.

= 1/[sqrt(x+0) + sqrt(x)
= 1/[sqrt(x) + sqrt(x)]
= 1/[2*sqrt(x)]
= (1/2)*x^(-1/2)

The expression above is the well-known derivative of sqrt(x), completing the proof.

## Using the Derivative

Derivatives are used to find tangent lines, tangent curves, instantaeous rates of change, and solve optimization problems among many things. For example, consider the curves defined by the functions f(x) = sqrt(x) and g(x) = ln(x). At what value of x do these two curves have the same instantaneous rate of change? In equivalent terms, at what value of x do their respective tangent lines have the same slope? To answer this question, we need to know that the derivative of sqrt(x) is 1/[2*sqrt(x)], and the derivative of ln(x) is 1/x. To find the value of x, we set the two expressions equal to each other and solve for x:

1/[2*sqrt(x)] = 1/x
2*sqrt(x) = x
4x = x^2
4 = x

Therefore, at x = 4, both curves have a tangent line slope equal to 1/4. The graph below shows the two functions and their parallel tangent lines at x = 4.

## Another Example

Let B(n) be an integer sequence defined as the y-intercept of the tangent on the curve f(x) = sqrt(x) at x = n, for n ≥ 1. In other words, for every positive integer n, consider the tangent line on the graph of y = sqrt(x) at x = n. B(n) is where this tangent line crosses the y-axis. What is the formula for B(n) in terms of n?

To solve this problem, let's find the slope of the tangent line at x = n. Using the derivative formula, we know it is 1/[2*sqrt(n)]. The tangent line also passes through the point (n, sqrt(n)) which is where it is tangent to the graph of sqrt(x). And with a slope and a point, you can find the "y = mx + b" form of the tangent line:

y - sqrt(n) = (1/[2*sqrt(n)])(x - n)
y - sqrt(n) = x/[2*sqrt(n)] - sqrt(n)/2
y = x/[2*sqrt(n)] + sqrt(n)/2

The y-intercept is sqrt(n)/2, therefore B(n) = sqrt(n)/2.

## Derivative of x^(1/3) Cube Root of x

The derivative of the cube root of x, cbrt(x) = x^(1/3) can also be found using limits in a process similar to that above. When you simplify the expression

im(h→0) [(x+h)^(1/3) - x^(1/3)]/h

you start by multiplying the top and bottom of the fraction by (x+h)^(2/3) + (x(x+h))^(1/3) + x^(2/3), which gives you

lim(h→0) [(x+h) - (x)] / [h*((x+h)^(2/3) + (x(x+h))^(1/3) + x^(2/3))]

= lim(h→0) [h] / [h*((x+h)^(2/3) + (x(x+h))^(1/3) + x^(2/3))]

= lim(h→0) 1 / [(x+h)^(2/3) + (x(x+h))^(1/3) + x^(2/3)]

= 1 / [x^(2/3) + x^(2/3) + x^(2/3)]

= 1/[3*x^(2/3)]

= (1/3)x^(-2/3)

In the same way you can show that the derivative of the nth root function f(x) = x^(1/n) is always f'(x) = (1/n)x^[(1-n)/n].

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