How do you differentiate the square root of a linear expression like the square root of 3x plus 2?

Use the chain rule. Let u equal 3x plus 2, so y equals u to the power one half. Then the derivative is one half times u to the power negative one half times the derivative of u, which is 3. This gives 3 over 2 times the square root of 3x plus 2.

How do you find the gradient of a tangent to a curve like y equals root x?

The gradient of the tangent at a point is the value of the derivative there. For y equals the square root of x, the derivative is 1 over 2 root x, so at x equals 9 the gradient is 1 over 2 times 3, which is one sixth.

How do you differentiate the square root of a linear expression like the square root of 3x plus 2?

Use the chain rule. Let u equal 3x plus 2, so y equals u to the power one half. Then the derivative is one half times u to the power negative one half times the derivative of u, which is 3. This gives 3 over 2 times the square root of 3x plus 2.

How do you find the gradient of a tangent to a curve like y equals root x?

The gradient of the tangent at a point is the value of the derivative there. For y equals the square root of x, the derivative is 1 over 2 root x, so at x equals 9 the gradient is 1 over 2 times 3, which is one sixth.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Differentiating rational powers

Q: How do you differentiate a function with a rational power?

First rewrite it as a single power of x, turning surds into fractional indices and reciprocals into negative indices, for example the square root of x becomes x to the power one half. Then apply the power rule: multiply by the index and subtract one from it. So the derivative of x to the power one half is one half times x to the power negative one half, which is 1 over 2 root x.

Q: How do you differentiate the square root of x?

Write the square root of x as x to the power one half. The power rule gives one half times x to the power negative one half, which equals 1 divided by 2 times the square root of x.

Q: How do you differentiate 1 over x squared?

Rewrite 1 over x squared as x to the power negative two. The power rule gives negative two times x to the power negative three, which is negative 2 over x cubed. The derivative is negative because the index is negative.

Q: Why do you rewrite a surd before differentiating?

The power rule only applies to a power written as x to the power n. A surd such as the square root of x or a reciprocal such as 1 over x squared is not in that form, so you must first rewrite it as x to the power one half or x to the power negative two before the power rule can be used.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a rational power is a power with a negative or fractional index. To differentiate one, first rewrite every surd or reciprocal as a single power of \(x\), then apply the power rule \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) — extended here to sums of such terms, to the chain rule for powers of linear and quadratic expressions, and to tangents, normals and rates of change.

A rational power of \(x\) is a term \(x^{n}\) whose index \(n\) is a fraction or a negative number, such as \(x^{1/2}=\sqrt{x}\) or \(x^{-2}=\dfrac{1}{x^{2}}\). Differentiating one uses the same power rule as for whole-number powers — but you must first write the function in the form \(x^{n}\).

The power rule holds for every rational \(n\):

\(\dfrac{d}{dx}x^{n}=n\,x^{n-1}.\)

So the whole skill is the rewrite: convert surds to fractional indices (\(\sqrt{x}=x^{1/2}\), \(x\sqrt{x}=x^{3/2}\)) and reciprocals to negative indices (\(\dfrac{1}{x^{2}}=x^{-2}\)), differentiate, then convert back. When the power is of a linear or quadratic expression, such as \(\sqrt{3x+2}\), the chain rule supplies the extra inner-derivative factor.

Key idea. Rewrite as \(x^{n}\), then \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) (any rational \(n\)). For \((ax+b)^{n}\), the chain rule gives \(n\,(ax+b)^{n-1}\times a\).

Gradient of the tangent \(=\dfrac{1}{2\sqrt{x}}\); at \((4,2)\) it is \(\dfrac{1}{4}\)

The tangent gradient falls as \(x\) grows: at \((1,1)\) it is \(\dfrac{1}{2}\)

The power rule, valid for every rational index \(n\):

\[\dfrac{d}{dx}x^{n}=n\,x^{n-1}\]

\frac{d}{d x} x^{n} = n x^{n - 1}

The rewrites that put a surd or reciprocal into that form:

\[\sqrt{x}=x^{1/2},\quad \sqrt[3]{x}=x^{1/3},\quad x\sqrt{x}=x^{3/2},\quad \dfrac{1}{x^{2}}=x^{-2},\quad \dfrac{1}{\sqrt{x}}=x^{-1/2}\]

\sqrt{x} = x^{1 / 2}

For a power of a linear or quadratic expression, the chain rule:

\[\dfrac{d}{dx}\big(ax+b\big)^{n}=n\,(ax+b)^{n-1}\times a\]

n (a x + b)^{n - 1} \times a

Tangents and normals. The gradient of the tangent at a point is the value of the derivative there. The normal is perpendicular, so its gradient is \(m_{\text{n}}=-\dfrac{1}{m}\).

How to differentiate a rational power

Rewrite in index form. Turn every surd into a fractional index and every reciprocal into a negative index (e.g. \(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{x^{2}}=x^{-2}\)).
Apply the power rule. Multiply by the index and subtract \(1\) from it: \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\). Differentiate a sum term by term.
Use the chain rule if needed. For \((ax+b)^{n}\), multiply by the inner derivative: \(n\,(ax+b)^{n-1}\times a\).
Convert back and apply. Rewrite the answer in surd or reciprocal form, then substitute a value for a gradient, tangent, normal or rate of change.

Rates of change. If a quantity varies as a root of time, such as \(d=5\sqrt{t}\), its rate of change is the derivative \(\dfrac{dd}{dt}=\dfrac{5}{2\sqrt{t}}\) — the instantaneous speed at time \(t\).

Example 1 — Fractional index

Differentiate \(y=\sqrt{x}\).

Solution

Rewrite as \(x^{1/2}\), then apply the power rule.

\(y\)	\(=\)	\(x^{1/2}\)
\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{2}x^{-1/2}\)
	\(=\)	\(\dfrac{1}{2\sqrt{x}}\)

\frac{d y}{d x} = \frac{1}{2 \sqrt{x}}

Example 2 — Negative index

Differentiate \(y=\dfrac{1}{x^{2}}\).

Solution

Write as \(x^{-2}\) and keep the negative sign.

\(y\)	\(=\)	\(x^{-2}\)
\(\dfrac{dy}{dx}\)	\(=\)	\(-2x^{-3}\)
	\(=\)	\(-\dfrac{2}{x^{3}}\)

\frac{d y}{d x} = - \frac{2}{x^{3}}

Example 3 — Chain rule

Differentiate \(y=\sqrt{3x+2}\).

Solution

Let \(u=3x+2\), so \(y=u^{1/2}\); multiply by \(u'=3\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{2}u^{-1/2}\times 3\)
	\(=\)	\(\dfrac{3}{2\sqrt{3x+2}}\)

\frac{d y}{d x} = \frac{3}{2 \sqrt{3 x + 2}}

Example 4 — Tangent gradient

Find the gradient of the tangent to \(y=\sqrt{x}\) at \(x=9\).

Solution

The gradient is the derivative evaluated at \(x=9\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{2\sqrt{x}}\)
\(\left.\dfrac{dy}{dx}\right\|_{x=9}\)	\(=\)	\(\dfrac{1}{2\sqrt{9}}=\dfrac{1}{6}\)

m = \frac{1}{6}

Common pitfalls

Rewrite the surd first. The power rule applies only to \(x^{n}\); you cannot differentiate \(\sqrt{x}\) or \(\dfrac{1}{x^{2}}\) as written — convert to \(x^{1/2}\) or \(x^{-2}\) first.

Subtract \(1\) from the index. The new index is \(n-1\): \(\dfrac{d}{dx}x^{1/2}=\dfrac{1}{2}x^{-1/2}\), not \(\dfrac{1}{2}x^{1/2}\).

Keep the sign of a negative index. \(\dfrac{d}{dx}x^{-2}=-2x^{-3}\); dropping the minus sign gives the wrong sign for the derivative of \(\dfrac{1}{x^{2}}\).

Do not forget the chain-rule factor. For \((ax+b)^{n}\) you must multiply by the inner derivative \(a\); leaving it out is the most common slip in \(\sqrt{3x+2}\)-type questions.

Frequently asked questions

How do you differentiate a function with a rational power?

Rewrite it as \(x^{n}\) — surds become fractional indices, reciprocals become negative indices — then apply \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\).

How do you differentiate the square root of x?

Write \(\sqrt{x}=x^{1/2}\); then \(\dfrac{dy}{dx}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}\).

How do you differentiate 1 over x squared?

Write \(\dfrac{1}{x^{2}}=x^{-2}\); then \(\dfrac{dy}{dx}=-2x^{-3}=-\dfrac{2}{x^{3}}\). The derivative is negative.

How do you differentiate the square root of 3x plus 2?

Use the chain rule with \(u=3x+2\): \(\dfrac{dy}{dx}=\dfrac{1}{2}u^{-1/2}\times 3=\dfrac{3}{2\sqrt{3x+2}}\).

Why do you rewrite a surd before differentiating?

Because the power rule only applies to a power \(x^{n}\); a surd or reciprocal must first be written in that form.

How do you find the gradient of a tangent to y = root x?

Evaluate the derivative at the point. For \(y=\sqrt{x}\), \(\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\), so at \(x=9\) the gradient is \(\dfrac{1}{6}\).

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