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Derivative of Square Root of 3x: Formula, 3 Methods & Solved Examples

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Derivative of Square Root of 3x: Formula, 3 Methods & Solved Examples

You’re working through a calculus problem, and there it is: √(3x). A square root with a 3 tucked inside. Your pen hovers. Do you treat the 3 like a coefficient? Does it change the rule? Where does it go?

Take a breath. This one is friendlier than it looks, and once you see how it works, you’ll be able to handle a whole family of similar problems.

In this guide, we’ll go through the formula, three different ways to get there, a handful of solved examples, and the slip-ups that quietly cost students marks on exams.

The derivative of the square root of 3x is:

d/dx [√(3x)] = 3 / (2√(3x))

You’ll also see it written as √3 / (2√x), or as ½·√(3/x). Don’t panic if your textbook shows a different-looking version. They’re all the same thing wearing different outfits, and we’ll prove that in a minute. The formula works for every x greater than 0.

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☰ Table of Contents

    Let’s Look at What We’re Dealing With

    Calculus rules like numbers more than they like roots. So the trick to differentiating almost any radical function is to turn the root into a power:

    √(3x) = (3x)^(1/2)

    That’s it. That one move unlocks the power rule and the chain rule. A few reminders before we start, because we’ll lean on them all the way through:

    • Power rule: d/dx [xⁿ] = n·xⁿ⁻¹
    • Chain rule: differentiate the outside, then multiply by the derivative of the inside
    • Constant multiple rule: a constant in front just tags along

    Where does this derivative actually exist?

    The function √(3x) is defined for x ≥ 0. But its derivative only exists for x > 0. At x = 0, the formula puts a zero in the denominator, and that’s a no-go. If you sketch the graph, you’ll see why: right at the origin the curve shoots straight up, so the tangent line is vertical, and a vertical line doesn’t have a regular slope.

    It’s a small detail, but exam questions love to ask about it.

    Method 1: The Chain Rule

    This is the method most teachers want to see, and it’s the one worth mastering, because it works on far messier functions later.

    Start by writing y = (3x)^(1/2). Think of it as two layers. The outer layer is “something to the power of ½,” and the inner layer is 3x.

    Differentiate the outer layer first, leaving the inside alone: (1/2)·(3x)^(−1/2).

    Then multiply by the derivative of the inside. The derivative of 3x is just 3.

    Put it together:

    dy/dx = (1/2)·(3x)^(−1/2)·3 = 3 / (2√(3x))

    Notice that the 3 on top comes from that last step. This is the piece people forget, so keep an eye on it.

    Method 2: The Power Rule Shortcut

    Here’s a little trick I like because it avoids the chain rule altogether. Square roots split nicely over multiplication, so:

    √(3x) = √3 · √x

    And √3 is just a number, about 1.732. A constant. So we can pull it out front and differentiate √x the way we always do:

    d/dx [√3 · x^(1/2)] = √3 · (1/2)·x^(−1/2) = √3 / (2√x)

    Wait, is that really the same as 3/(2√(3x))? Check it. The denominator 2√(3x) is 2·√3·√x, so we have 3 / (2·√3·√x). Since 3 = √3·√3, one √3 cancels with the one on the bottom, leaving √3 / (2√x). Same answer.

    This shortcut is great when the inside is just a constant times x. It’s less useful when the inside gets more complicated, like 3x + 1.

    Method 3: First Principles (The “Why It Works” Method)

    If your teacher says “prove it from the definition,” this is what they mean. It’s longer, but it’s also the most satisfying, because nothing is taken on trust.

    The definition is:

    f′(x) = lim (h→0) [f(x + h) − f(x)] / h

    For f(x) = √(3x), that becomes:

    lim (h→0) [√(3(x + h)) − √(3x)] / h

    If you plug in h = 0 right away, you get 0/0, which tells you nothing. So we use a classic move: multiply the top and bottom by the conjugate, √(3(x + h)) + √(3x).

    The top turns into a difference of squares:

    3(x + h) − 3x = 3h

    So now we have:

    lim (h→0) 3h / { h · [√(3(x + h)) + √(3x)] }

    The h on top and bottom cancels, which is exactly what we were hoping for:

    lim (h→0) 3 / [√(3(x + h)) + √(3x)]

    Now let h shrink to zero, and both square roots become √(3x):

    3 / (2√(3x))

    Three different roads, one destination. That’s always a reassuring sign in math.

    Which Method Should You Use?

    Honestly, it depends on the moment.

    If you’re doing homework or an exam and just need the answer, the chain rule is your best friend, because it scales up to anything. If the function is as simple as √(3x), the power rule shortcut is the quickest. And if the question says “from first principles” or “prove,” you don’t have a choice.

    If you only have time to really learn one, learn the chain rule.

    The Second Derivative

    Sometimes you need to go one step further, for concavity or for motion problems. Start with the simplified first derivative:

    f′(x) = (√3 / 2)·x^(−1/2)

    Differentiate once more:

    f″(x) = (√3 / 2)·(−1/2)·x^(−3/2) = −√3 / (4x^(3/2))

    You may also see this as −9 / [4(3x)^(3/2)]. Same thing.

    What does the negative sign tell us? The graph of √(3x) is concave down for all x > 0. It’s still climbing, but the climb keeps getting gentler, like a hill that flattens out the farther you walk.

    Solved Examples

    Example 1: Finding a tangent line

    Find the slope of y = √(3x) at x = 12, and write the tangent line.

    The derivative is 3/(2√(3x)). At x = 12, the inside is 36, and √36 = 6, so the slope is 3/12 = 1/4.

    The point on the curve is (12, 6). The tangent line is y − 6 = ¼(x − 12), which simplifies to y = x/4 + 3.

    Example 2: When it shows up in a product

    Differentiate f(x) = x²·√(3x).

    You could use the product rule: 2x·√(3x) + x²·3/(2√(3x)). That’s correct, but it’s a bit clunky to simplify.

    Here’s the cleaner route. Rewrite the function first: x²·√3·x^(1/2) = √3·x^(5/2). Then the power rule gives:

    f′(x) = (5√3 / 2)·x^(3/2)

    Both approaches give the same value. The lesson is worth remembering: rewriting before you differentiate can save you a lot of work.

    Example 3: When it shows up in a quotient

    Differentiate g(x) = √(3x) / x.

    With the quotient rule:

    g′(x) = [ (3/(2√(3x)))·x − √(3x) ] / x²

    Here’s a nice simplification: 3x / (2√(3x)) is the same as √(3x)/2. So the top becomes √(3x)/2 − √(3x) = −√(3x)/2, and:

    g′(x) = −√(3x) / (2x²) = −√3 / (2x^(3/2))

    Want to double-check? Rewrite g(x) as √3·x^(−1/2) and use the power rule. You’ll land in the same place.

    Example 4: A real-world rate of change

    Water fills a container so that its height after t seconds is h(t) = √(3t) centimeters. How fast is the water level rising at t = 27 seconds?

    The rate is h′(t) = 3 / (2√(3t)). At t = 27, we get √81 = 9, so:

    h′(27) = 3 / 18 = 1/6 cm per second (about 0.17)

    At that moment the water was 9 cm deep and still rising, just slowly. That matches what the negative second derivative told us earlier: the rate keeps dropping.

    Example 5: A close cousin, √(3x + 1)

    Differentiate y = √(3x + 1) and find the slope at x = 1.

    The chain rule applies the same way. The outside gives 1/(2√(3x + 1)), and the inside, 3x + 1, has derivative 3. So:

    y′ = 3 / (2√(3x + 1))

    At x = 1, that’s 3 / (2·2) = 3/4.

    Same pattern, slightly bigger inside. Once you spot it, a whole group of problems gets easier.

    Example 6: Second derivative at a point

    Find f″(3) for f(x) = √(3x).

    Using f″(x) = −√3 / (4x^(3/2)):

    f″(3) = −√3 / (4·3√3) = −1/12

    Mistakes That Catch Almost Everyone

    Let’s be honest about where things go wrong. These come up again and again.

    Forgetting the 3 on top. Writing 1/(2√(3x)) instead of 3/(2√(3x)) is the classic. The chain rule asks you to multiply by the derivative of the inside, and the inside here is 3x, whose derivative is 3.

    Dropping the 3 completely. Some students end up with 1/(2√x), as though the 3 never existed. A constant inside the root absolutely changes the answer.

    Skipping steps. It feels slow to write out every line, but jumping ahead is exactly how small sign and coefficient errors sneak in. Write it out and check whether your final form matches the answer key.

    Forgetting the domain. If a question asks about x = 0 or a negative x, the honest answer is that the derivative doesn’t exist there.

    Mixing up √(3x) with 3√x. They look similar, but they aren’t equal. √(3x) = √3·√x, while 3√x has a different constant out front. The derivative of 3√x is 3/(2√x), which is not the same as √3/(2√x).

    Why bother? Where This Shows Up

    Square root functions describe lots of things that grow quickly at first and then ease off: the height of liquid in certain containers, the speed of a falling object compared to the distance it’s fallen, and some cost or output curves where each extra unit gives a bit less than the last. The derivative tells you how fast the quantity is changing right now, which is the whole point of calculus, and why physics, engineering, and economics courses keep coming back to it.

    Quick Revision Sheet

    If you’re short on time, here’s the essence:

    • √(3x) = (3x)^(1/2)
    • First derivative: 3 / (2√(3x)), also √3 / (2√x)
    • Second derivative: −√3 / (4x^(3/2))
    • The derivative exists only for x > 0
    • Tools to know: chain rule, power rule, first principles

    Conclusion

    The derivative of the square root of 3x really comes down to two habits: rewrite the root as a power, and never forget the derivative of what’s inside. Try a few variations on your own, like √(3x + 5), √(3x²), or a product with x³, and you’ll find the pattern sticks faster than you’d expect. And if you ever doubt an answer, run it through a second method. When two routes agree, you can trust the result.

    Frequently Asked Questions

    It’s 3 / (2√(3x)), which can also be written as √3 / (2√x).

    Rewrite √(3x) as √3·x^(1/2), pull out √3, and differentiate x^(1/2) to get √3 / (2√x).

    No. The function itself is defined there, but the slope becomes infinite, so the derivative doesn’t exist.

    It comes from the chain rule. The derivative of the inner function 3x is 3, and it multiplies the outer derivative.

    Not quite. √(3/x) = √3·x^(−1/2), and its derivative is −√3 / (2x^(3/2)). Notice the negative sign and the different power.

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