Derivative of Square Root of `x`
Introduction
To understand the derivative of the square root of `x`, let's break it down. Imagine a function where one variable `y` depends on another variable, `x`. The derivative of this function tells how much `y` changes as `x` changes. In calculus, this process is called differentiation. Now, when we say square root of `x`, we're talking about the mathematical operation where we find the number which, when multiplied by itself, gives you `x`. In symbols, we write this as `sqrtx` or `x^(1/2)`.
The derivative of the function \(f(x) = \sqrt{x}\) is a fundamental concept in calculus that captures how the function changes at any given point on its domain. The square root function is particularly interesting because it is defined for all real numbers greater than or equal to zero. Square root function is not defined for negative numbers as square root of a negative number is a complex number and not a real number.
What Is the Derivative of the Square Root of `x`?
The derivative of the square root of `x` tells us how the function \(f(x) = \sqrt{x}\) changes as `x` changes. This derivative is found using rules in calculus. We use `d/dx(sqrtx)` (or) `(sqrtx)'` to indicate the derivative of `sqrt(x)` with respect to `x`.
Now, let's get into how we find the derivative of the square root of `x`. There are different methods for this, but one common way is using the power rule. This rule says that if you have a function of `x` raised to some power, then the derivative of that function is `n` times `x` raised to the power `(n-1)`. For the square root of `x`, which is `x^(1/2)`, we can apply this rule by setting `n = 1/2`. When we do this, we find that the derivative is `(1/2)` times `x` raised to the power `(-1/2)`, written as `\frac{1}{2}x^{\frac{-1}{2}}` or more simply, `(1/(2sqrtx))`.
Consequently,
If `(sqrt (x))' = \frac{1}{2}x^{\frac{-1}{2}}`, then `d/dx (sqrt (x)) = \frac{1}{2}x^{\frac{-1}{2}}`
For example, if we want to find the rate of change of root `x`, we use the power rule and the first principle. This helps us understand how fast the square root function changes as `x` changes. We can also use it to find the slopes of tangent lines and the velocity of moving objects.
Formula for the Derivative of Root `x`
The formula for finding the derivative of the square root of `x` is pretty straightforward. It's expressed as
`\frac{d(\sqrt{x})}{dx}` or `(\sqrt{x})' = \frac{1}{2}x^{-\frac{1}{2}}`, or `\frac{1}{2\sqrt{x}}`.
We can prove that the derivative of `sqrtx` is `\frac{1}{2}x^{-\frac{1}{2}}` or `\frac{1}{2\sqrt{x}}` using various ways. A couple of ways are listed below:
`1`. Using the First Principle of Derivatives, which involves calculating the limit of the difference quotient as the interval approaches zero.
`2`. Using the Power Rule of Differentiation, a shortcut method applicable for functions involving powers of `x`.
Proving Derivative of Square Root of `x` Using First Principle
We start with the definition of the First Principle which says
`f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}`
Using this definition, we want to prove the derivative of \( \sqrt{x} \). So, let \( f(x) = \sqrt{x} \).
`f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}`
Rationalize the numerator by multiplying and dividing by \( \sqrt{x+h} + \sqrt{x} \):
`f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}`
`= \lim_{h \to 0} \frac{(x+h - x)}{h(\sqrt{x+h} + \sqrt{x})}`
`= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}`
`= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}`
`= \frac{1}{\sqrt{x} + \sqrt{x}}`
`= \frac{1}{2\sqrt{x}}`
So, the derivative of \( \sqrt{x} \) using the first principles is `\frac{1}{2\sqrt{x}}`.
Note that `\frac{1}{2\sqrt{x}}` can also be written as `\frac{1}{2}x^{-1/2}`.
Proving Derivative of Square Root of `x` Using Power Rule
Alternatively, we can use the Power Rule for differentiation to prove the derivative of square root of `x`. According to this rule, the derivative of `x^n` is `nx^(n-1)`.
Since the square root of `x` can be written as `x^(1/2)`, we apply the Power Rule by substituting `n = 1/2`. This gives us `(1/2)x^(1/2 - 1)`, which simplifies to `1/(2sqrtx)`.
Thus, both methods yield the same result: the derivative of the square root of `x` is `1/(2sqrtx)`.
Uses of the Derivative of Root `x`
One use of the derivative of `sqrtx` is to find the slope of square root functions. When you want to find how fast something changes at a certain point, you need to find its derivative.
So, if you have a function like `sqrt(2x + 5)`, and you want to know how it's changing, you use the chain rule and the derivative of `sqrtx`.
Example: Find the derivative of \( \sqrt{x^2 + 1} \).
Solution:
Given function: \( y = \sqrt{x^2 + 1} \)
Step `1`. Apply the chain rule:
`\frac{d}{dx}\sqrt{x^2 + 1} = \frac{1}{2\sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x^2 + 1)`
Step `2`. Find `\frac{d}{dx}(x^2 + 1)`:
`\frac{d}{dx}(x^2 + 1) = 2x`
Step `3`. Substitute `\frac{d}{dx}(x^2 + 1)` into the chain rule:
`\frac{d}{dx}\sqrt{x^2 + 1} = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x`
`= \frac{x}{\sqrt{x^2 + 1}}`
So, the derivative of \( \sqrt{x^2 + 1} \) with respect to \( x \) is `\frac{x}{\sqrt{x^2 + 1}}`.
Solved Examples
Example `1`: Find the derivative of `sqrt(4x^2 + 1)`.
Solution:
To find the derivative of the function \( f(x) = \sqrt{4x^2 + 1} \), we use the chain rule. The chain rule states that if \( f(x) = \sqrt{g(x)} \), then the derivative \( f'(x) \) is given by:
`f'(x) = \frac{1}{2\sqrt{g(x)}} g'(x)`
Here, \( g(x) = 4x^2 + 1 \). First, find the derivative \( g'(x) \):
`g'(x) = \frac{d}{dx}(4x^2 + 1) = 8x`
Now, apply the chain rule:
`f'(x) = \frac{1}{2\sqrt{4x^2 + 1}} \cdot 8x`
Simplifying, we have:
`f'(x) = \frac{4x}{\sqrt{4x^2 + 1}}`
Therefore, the derivative of the function \( f(x) = \sqrt{4x^2 + 1} \) is `\frac{4x}{\sqrt{4x^2 + 1}}`.
Example `2`: Find the derivative of `x \sqrt{\tan x}`.
Solution:
To differentiate the function \( f(x) = x \sqrt{\tan x} \), we use the product rule. The product rule states that if \( f(x) = g(x) \cdot h(x) \), then:
\( f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) \)
For \( f(x) = x \sqrt{\tan x} \), let:
\( g(x) = x \quad \text{and} \quad h(x) = \sqrt{\tan x} \)
The derivatives of \( g(x) \) and \( h(x) \) are:
\( g'(x) = 1 \)
`h(x) = (\tan x)^{1/2}`
Applying the chain rule to differentiate \( h(x) \):
`h'(x) = \frac{1}{2} (\tan x)^{-1/2} \cdot \sec^2 x`
This follows from the derivative of \( \tan x \), which is \( \sec^2 x \).
Now, substitute these derivatives into the product rule:
`f'(x) = 1 \cdot \sqrt{\tan x} + x \cdot \frac{1}{2} (\tan x)^{-1/2} \cdot \sec^2 x`
Simplifying, we obtain:
`f'(x) = \sqrt{\tan x} + \frac{x \sec^2 x}{2 \sqrt{\tan x}}`
Therefore, the derivative \( f'(x) \) of the function \( f(x) = x \sqrt{\tan x} \) is:
`f'(x) = \sqrt{\tan x} + \frac{x \sec^2 x}{2 \sqrt{\tan x}}`
`= \frac{2\tan(x) + x\sec^2(x)}{2 \sqrt{\tan x}}`
Example `3`: Find the derivative of `sqrt(5sin(x))`.
Solution:
The function is given by:
\( f(x) = \sqrt{5 \sin x} \)
We can rewrite this function as:
`f(x) = (5 \sin x)^{1/2}`
To differentiate \( f(x) \), apply the chain rule. Let \( u = 5 \sin x \), then `f(u) = u^{1/2}`. The derivatives are:
`\frac{df}{du} = \frac{1}{2} u^{-1/2}`
`\frac{du}{dx} = 5 \cos x`
Applying the chain rule:
`\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{1}{2} (5 \sin x)^{-1/2} \cdot 5 \cos x`
Simplifying the expression, we get:
`\frac{df}{dx} = \frac{5}{2} \cdot \frac{\cos x}{\sqrt{5 \sin x}}`
Thus, the derivative of \( f(x) = \sqrt{5 \sin x} \) is:
`\frac{df}{dx} = \frac{5 \cos x}{2 \sqrt{5 \sin x}}`
Example `4`: What is the derivative of `x sqrtlog x`?
Solution:
To find the derivative of \( x \sqrt{\log{x}} \), we'll use the product rule and the chain rule.
Given function: \( y = x \sqrt{\log{x}} \)
Let's break it down step by step:
`1`. Product Rule:
\( f(x) = x \)
\( g(x) = \sqrt{\log{x}} \)
\( f'(x) = 1 \)
`g'(x) = \frac{1}{2\sqrt{\logx}} \cdot \frac{1}{x}`
`2`. Apply the Product Rule:
\( y' = f'(x) \cdot g(x) + f(x) \cdot g'(x) \)
`y' = 1 \cdot \sqrt{\logx} + x \cdot \frac{1}{2\sqrt{\logx}} \cdot \frac{1}{x}`
`3`. Simplify:
`y' = \sqrt{\logx} + \frac{1}{2\sqrt{\logx}}`
`4`. Combine fractions:
`y' = \frac{2\logx+1}{2\sqrt{\logx}}`
So, the derivative of \( x \sqrt{\log{x}} \) with respect to \( x \) is `\frac{2\logx+1}{2\sqrt{\logx}}`.
Example `5`: Find the derivative of `sqrt(2/x)`.
Solution:
To find the derivative of the function `f(x) = \sqrt{\frac{2}{x}}`, we can use the chain rule. First, we'll express the function using a simpler notation for differentiation:
Let:
`u = \frac{2}{x}`
Then:
`f(x) = \sqrt{u} = u^{1/2}`
Now, apply the chain rule. We need to find \( u'(x) \) first, which is the derivative of `\frac{2}{x}`. Using the quotient rule or power rule, we find:
`u'(x) = \frac{d}{dx} \left(\frac{2}{x}\right) = -\frac{2}{x^2}`
The derivative of `f(u) = u^{1/2}` with respect to \( u \) is:
`f'(u) = \frac{1}{2} u^{-1/2}`
Now, using the chain rule, the derivative \( f'(x) \) is:
`f'(x) = f'(u) \cdot u'(x) = \frac{1}{2} u^{-1/2} \cdot \left(-\frac{2}{x^2}\right)`
Simplify by substituting `u = \frac{2}{x}` back into the expression:
`f'(x) = \frac{1}{2} \left(\frac{2}{x}\right)^{-1/2} \cdot \left(-\frac{2}{x^2}\right) = \frac{1}{2} \left(\frac{\sqrt{x}}{\sqrt{2}}\right) \cdot \left(-\frac{2}{x^2}\right)`
Further simplifying:
`f'(x) = \frac{1}{2} \cdot \frac{\sqrt{x}}{\sqrt{2}} \cdot \left(-\frac{2}{x^2}\right) = -\frac{1}{\sqrt{2}} \cdot \frac{1}{x\sqrt{x}}`
Finally, this can be rewritten in a slightly more simplified form:
`f'(x) = -\frac{1}{x\sqrt{2x}}`
Thus, the derivative of `f(x) = \sqrt{\frac{2}{x}}` is:
`f'(x) = -\frac{1}{x\sqrt{2x}}`
Practice Problems
Q`1`. Find the derivative of `sqrt(3x + 2)` with respect to `x`.
- `1/(2sqrt(3x + 2))`
- `3/(2sqrt(3x + 2))`
- `1/(6sqrt(3x + 2))`
- `2/(3sqrt(3x + 2))`
Answer: b
Q`2`. Determine the derivative of `sqrt(5x - 4)`.
- `1/(2sqrt(5x - 4))`
- `5/(2sqrt(5x - 4))`
- `1/(10sqrt(5x - 4))`
- `2/(5sqrt(5x - 4))`
Answer: b
Q`3`. What is the derivative of `sqrt(5x^2 - 6)`?
- `1/(2sqrt(2x - 7))`
- `(5 x)/sqrt(5 x^2 - 6)`
- `1/(4sqrt(2x - 7))`
- `3/(2sqrt(2x - 7))`
Answer: b
Q`4`. What is the derivative of `(sqrtx +4)\sin x`?
- \(\sqrt{x}\cos(x) + 4\sin(x) \)
- \(\cos(x)\sqrt{x} + 4\sin(x) \)
- `\frac{1}{2\sqrt{x}}\sin(x) + 4\cos(x)`
- `\frac{\sin(x) + 2(x+4)\cos(x)}{2\sqrt{x+4}}`
Answer: d
Q`5`. Find the derivative of `logsqrtx`.
- `1/(2x)`
- `3/(2sqrt(6x))`
- `1/12`
- `2/(5sqrt5)`
Answer: a
Frequently Asked Questions
Q`1`. What functions cannot be differentiated?
Answer: Some functions can't be smoothly "figured out" at certain points. Imagine drawing the graph of a function, and suddenly it hits a point where it goes straight up like a wall. That's called a vertical tangent line. When this happens, the function isn't differentiable at that point. It's like calculating how steep a wall is at a certain spot—it's impossible because the slope is undefined. So, if the graph has a spot where it goes straight up, we say the function isn't differentiable there.
Q`2`. What is the derivative of `sqrt2`?
Answer: The derivative of `sqrt2` is zero. It is because `sqrt2` is a constant number and in calculus, the derivative of a constant is always equal to zero.
Q`3`. What does the derivative of `sqrtx` represent?
Answer: The derivative of `sqrtx` represents the rate of change, or slope, of a square root function concerning its input variable `x`. It tells us how fast the square root function changes with respect to `x` at any given point.
Q`4`. How do you find the derivative of `sqrtx`?
Answer: To find the derivative of `sqrtx`, you use the power rule for differentiation, which states that the derivative of `x^n` is `n*x^(n-1)`. For `sqrtx`, you can rewrite it as `x^(1/2)`, then apply the power rule to get `(1/2)*x^(-1/2)`. Simplifying, you get `1/(2sqrtx)`.
Q`5`. What is the significance of the `1/2` in the derivative formula for `sqrtx`?
Answer: The `1/2` in the derivative formula for `sqrtx` represents the power to which `x` is raised `(1/2)` when differentiating. It indicates that the rate of change of the square root function is half the reciprocal of the square root of `x`. This fraction determines the slope of the function at any point.