The derivative of the square root of 2x is 1 / √(2x) for every x > 0. So in calculus, this means that if a function is written as f(x) = √(2x), its rate of change at any positive value of x is given by f'(x) = 1 / √(2x). This result is useful because square root functions appear often in algebra, physics, geometry, and real-world modeling, especially when relationships involve areas, distances, or growth that changes more slowly as values get larger.
Introduction
Finding the derivative of the square root of 2x is a common calculus problem because it combines two important ideas: working with radical functions and applying the chain rule. At first glance, the expression √(2x) may look slightly different from a simple square root like √x, but the method is straightforward once the function is rewritten in a more useful form.
The function can be written as:
f(x) = √(2x)
or, using exponent notation:
f(x) = (2x)^{1/2}
Once written this way, you can apply standard differentiation rules. The final answer is:
f'(x) = 1 / √(2x)
This derivative is defined for x > 0. At x = 0, the original function exists because √(2·0) = 0, but the derivative does not exist as a finite number because the graph has a vertical tangent there.
Understanding the Function
Before differentiating, it helps to understand what the function represents. The expression:
f(x) = √(2x)
means “take 2x, then take its square root.” The function is only real-valued when the inside of the square root is nonnegative:
2x ≥ 0
Dividing by 2 gives:
x ≥ 0
So, the domain of f(x) = √(2x) is:
x ≥ 0
Still, when finding the derivative, the domain becomes slightly smaller. The derivative exists only when x > 0, because at x = 0, the slope becomes infinitely steep.
Graphically, f(x) = √(2x) looks like the graph of √x, but it is horizontally compressed. The curve starts at the origin and rises slowly as x increases.
Method 1: Using the Chain Rule
The most common way to find the derivative is by using the chain rule. The chain rule is used when a function is made of an “outer function” and an “inner function.”
For:
f(x) = √(2x)
rewrite it as:
f(x) = (2x)^{1/2}
Now identify the parts:
- Outer function: u^{1/2}
- Inner function: u = 2x
The chain rule says:
d/dx [g(u)] = g'(u) · u'
First, differentiate the outer function:
d/du [u^{1/2}] = 1/2 u^{-1/2}
Then differentiate the inner function:
d/dx [2x] = 2
Now multiply them together:
f'(x) = 1/2 (2x)^{-1/2} · 2
The 1/2 and 2 cancel:
f'(x) = (2x)^{-1/2}
Rewrite the negative exponent as a fraction:
f'(x) = 1 / (2x)^{1/2}
Therefore:
f'(x) = 1 / √(2x)
This is the derivative of the square root of 2x Small thing, real impact. Nothing fancy..
Method 2: Using the Power Rule
You can also rewrite the function in a different way before differentiating. Since:
√(2x) = √2 · √x
you can write:
f(x) = √2 · x^{1/2}
Here, √2 is just a constant. The power rule says:
d/dx [x^n] = n x^{n-1}
Apply it to x^{1/2}:
f'(x) = √2 · 1/2 x^{-1/2}
Simplify:
f'(x) = √2 / (2√x)
This answer looks different, but it is equivalent to 1 / √(2x). To see why, multiply numerator and denominator by √2:
√2 / (2√x) = √2 / (√2 · √2 · √x)
Since 2 = √2 · √2, this becomes:
√2 / (√2 · √(2x)) = 1 / √(2x)
So both forms are correct:
f'(x) = √2 / (2√x)
or
f'(x) = 1 / √(2x)
The second form is usually cleaner because it matches the original expression.
Method 3: Using First Principles
The derivative can also be found using the definition of the derivative, also called first principles. The derivative of a function f(x) is:
f'(x) = limₕ→₀ [f(x + h) - f(x)] / h
For **f(x
For (f(x)=\sqrt{2x}) the definition of the derivative is
[ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} =\lim_{h\to 0}\frac{\sqrt{,2(x+h),}-\sqrt{,2x,}}{h}. ]
Rewrite the numerator by rationalising:
[ \frac{\sqrt{2x+2h}-\sqrt{2x}}{h} ;=; \frac{\bigl(\sqrt{2x+2h}-\sqrt{2x}\bigr)\bigl(\sqrt{2x+2h}+\sqrt{2x}\bigr)} {h\bigl(\sqrt{2x+2h}+\sqrt{2x}\bigr)}. ]
The product in the numerator simplifies to (2h):
[ \frac{2h}{h\bigl(\sqrt{2x+2h}+\sqrt{2x}\bigr)} ;=; \frac{2}{\sqrt{2x+2h}+\sqrt{2x}}. ]
Now let (h) approach (0). Both square‑root terms tend to (\sqrt{2x}), so the limit becomes
[ f'(x)=\frac{2}{2\sqrt{2x}}=\frac{1}{\sqrt{2x}};, ]
exactly the same result we obtained with the chain rule and the power rule.
The calculation confirms that the derivative exists for all (x>0); at (x=0) the limit diverges, reflecting the vertical tangent there Which is the point..
A quick look at the second derivative
Differentiating once more,
[ f''(x)=\frac{d}{dx}!\left(\frac{1}{\sqrt{2x}}\right) =-\frac{1}{2},(2x)^{-3/2}\cdot 2 =-\frac{1}{(2x)^{3/2}} =-\frac{1}{2\sqrt{2},x^{3/2}}. ]
The negative sign tells us that (f(x)=\sqrt{2x}) is concave down on its domain (x>0).
Tangent line example
At a point (x_0>0) the slope of the tangent is (f'(x_0)=1/\
Tangent line at a specific point
If we pick a particular value (x_0>0), the tangent line to the curve (y=\sqrt{2x}) at that point has the form
[ y-y_0 = f'(x_0),(x-x_0), \qquad y_0=f(x_0)=\sqrt{2x_0}. ]
Substituting the slope (f'(x_0)=1/\sqrt{2x_0}) gives
[ y-\sqrt{2x_0} =\frac{1}{\sqrt{2x_0}},(x-x_0), ]
or, after a little algebra,
[ y = \frac{x}{\sqrt{2x_0}} + \sqrt{2x_0}\Bigl(1-\frac{1}{2}\Bigr) = \frac{x}{\sqrt{2x_0}} + \frac{\sqrt{2x_0}}{2}. ]
This line will be horizontal only when (x_0) tends to infinity, confirming that the curve has no horizontal tangents on its finite domain.
Summary
We have examined the derivative of (f(x)=\sqrt{2x}) from three complementary viewpoints:
- Chain rule – a quick, textbook approach that immediately yields (f'(x)=1/\sqrt{2x}).
- Power rule – after rewriting the function as a constant times a power of (x), the same result emerges, illustrating the equivalence of the two algebraic forms.
- First principles – the definition of the derivative provides a rigorous verification that the slope at any point (x>0) is indeed (1/\sqrt{2x}).
The second derivative, (f''(x)=-1/(2\sqrt{2},x^{3/2})), shows that the function is concave downward everywhere on its domain. The tangent‑line formula derived above gives a concrete geometric interpretation of the slope.
In all three methods, the final answer is the same:
[ \boxed{,f'(x)=\frac{1}{\sqrt{2x}},}. ]
This consistency across different techniques reinforces our confidence in the result and demonstrates the power of calculus to connect algebraic manipulation, limit processes, and geometric intuition.
