1.7 - Differentiating powers | Cambridge A-Level Mathematics

Chord gradients explain what a derivative means; rules make it quick to find one. You will differentiate rational powers, rewrite roots and fractions before differentiating, and find second derivatives. Keeping the original domain in view will prevent an algebraic answer being used where the function is undefined.

The power rule

How can you find the gradient of $y=x^{3/2}$ without building a new chord table at every point? The same rule that gives $2x$ for $x^2$ works for rational powers, including fractions and negative numbers.

For positive integer powers, expanding a nearby value explains the pattern. For example, the chord gradient for $x^3$ at $x=a$ is $3a^2+3ah+h^2$ ; the terms containing $h$ vanish in the limit. The result is $3a^2$ . The rule below extends to rational exponents on intervals where the function is defined and differentiable.

The power rule

For a constant rational number $n$ ,

\frac{d}{dx}(x^n)=nx^{n-1}.

Use the rule where the real power and its derivative are defined. A constant function has derivative zero.

In the formula booklet.

Multiply by the old exponent, then subtract $1$ from that exponent. For instance, $x^5$ gives $5x^4$ , and $x=x^1$ gives $1$ . A constant such as $7$ has a horizontal graph, so its derivative is $0$ .

Worked example 1: a fractional power

Question. Differentiate $y=x^{3/2}$ for $x>0$ , and find the gradient at $x=4$ .

Plan. Apply the power rule, then substitute into the derivative.

\begin{aligned} \frac{dy}{dx}&=\frac32x^{3/2-1} &&\text{multiply and reduce the power}\\ &=\frac32x^{1/2}\\ \left.\frac{dy}{dx}\right|_{x=4}&=\frac32\sqrt4=3. \end{aligned}

Answer. $dy/dx=\frac32\sqrt{x}$ , with gradient $3$ at $x=4$ .

Check. The nearby chord from $x=4$ to $4.01$ has gradient $(4.01^{3/2}-8)/0.01\approx3.00187$ , close to $3$ .

Change just the exponent's sign. A negative exponent makes the function fall on the positive domain.

Worked example 2: a negative fractional power

Question. Differentiate $y=x^{-3/2}$ for $x>0$ , and find the gradient at $x=4$ .

Plan. Keep the negative coefficient and subtract $1$ from $-3/2$ .

\begin{aligned} \frac{dy}{dx}&=-\frac32x^{-3/2-1}\\ &=-\frac32x^{-5/2}\\ \left.\frac{dy}{dx}\right|_{x=4}&=-\frac32\times\frac1{32}=-\frac3{64}. \end{aligned}

Answer. $dy/dx=-3/(2x^{5/2})$ , with gradient $-3/64$ at $x=4$ .

Check. $y=1/x^{3/2}$ decreases as positive $x$ grows, agreeing with the negative derivative. The nearby chord to $4.01$ has gradient approximately $-0.046729$ , close to $-3/64=-0.046875$ .

The domain matters. A reciprocal power excludes $x=0$ . For $\sqrt{x}$ , the real function allows $x\ge0$ , but the formula $1/(2\sqrt{x})$ gives a finite derivative only for $x>0$ . At $0$ , the curve has a vertical direction, not a finite gradient. Odd roots can allow negative inputs: $d(x^{1/3})/dx=\frac13x^{-2/3}$ for $x\ne0$ .

Watch out

$-\frac32-1=-\frac52$ , not $-\frac12$ . Subtracting $1$ makes a negative exponent more negative. Write it with a common denominator if needed.

Apply that same change to a new negative fractional power.

Sums and multiples

A root or a fraction may hide a simple power. Rewrite it before applying the rule: $\sqrt{x}=x^{1/2}$ , $1/x^2=x^{-2}$ , and $1/\sqrt{x}=x^{-1/2}$ . Differentiate each term and keep any constant multiplier.

Why can terms be treated separately? A chord's total change in $3f(x)-2g(x)$ is three times the change in $f$ minus twice the change in $g$ . Taking the limiting gradient preserves that combination.

Sums and constant multiples

Where $f$ and $g$ are differentiable and $a,b$ are constants,

\frac{d}{dx}[af(x)+bg(x)]=af'(x)+bg'(x).

The derivative of a constant is zero.

Not stated separately in the formula booklet: learn these rules.

Keep rewriting and differentiation as separate lines. This makes the exponents visible before they change.

Worked example 3: a root in a sum

Question. Differentiate $f(x)=3x^3-5\sqrt{x}+8$ , for $x>0$ .

Plan. Rewrite the root as a power and differentiate term by term.

\begin{aligned} f(x)&=3x^3-5x^{1/2}+8\\ f'(x)&=3(3x^2)-5\left(\frac12x^{-1/2}\right)+0\\ &=9x^2-\frac5{2\sqrt{x}}. \end{aligned}

Answer. $f'(x)=9x^2-5/(2\sqrt{x})$ .

Check. Adding $8$ only moves the graph upwards, so it should not alter any gradient. Its disappearance is appropriate.

Move only the root into the denominator. This changes its exponent and therefore the sign of its derivative contribution.

Worked example 4: a reciprocal root in a sum

Question. Differentiate $g(x)=3x^3-5/\sqrt{x}+8$ , for $x>0$ .

Plan. Write the reciprocal root as $x^{-1/2}$ , then keep the two negative signs separate.

\begin{aligned} g(x)&=3x^3-5x^{-1/2}+8\\ g'(x)&=9x^2-5\left(-\frac12x^{-3/2}\right)\\ &=9x^2+\frac5{2x^{3/2}}. \end{aligned}

Answer. $g'(x)=9x^2+5/(2x^{3/2})$ .

Check. At $x=1$ , $g'(1)=11.5$ . The chord from $1$ to $1.001$ has gradient approximately $11.5071$ , consistent with this value.

Algebra may also remove an apparent quotient. For $x\ne0$ , $(x^3+4x)/x=x^2+4$ , so its derivative is $2x$ on the original domain. Cancelling does not put the excluded input $x=0$ back into the original function.

Watch out

$d(5x^3)/dx=15x^2$ , not $5x^2$ . The original coefficient multiplies the exponent. Also, differentiating a constant gives $0$ , not that constant.

Rewrite first, then differentiate.

Repeated differentiation

The first derivative is another function, so apply the same rules to it to find the second derivative. Differentiate the complete first derivative before substituting a particular value of $x$ . Substituting too soon turns it into a number and loses how it changes.

Evaluating a derivative

$f'(a)$ means find the gradient function $f'(x)$ , then evaluate it at $x=a$ . Similarly, $f''(a)$ means evaluate the second derivative there. These are different from $f(a)$ , the function's height.

Use one line for each differentiation so that each change of exponent can be checked.

Worked example 5: two polynomial derivatives

Question. For $y=2x^4-3x^2+7$ , find $dy/dx$ , $d^2y/dx^2$ , and both derivatives at $x=1$ .

Plan. Differentiate twice, then substitute $x=1$ into the two resulting functions.

\begin{aligned} \frac{dy}{dx}&=8x^3-6x\\ \frac{d^2y}{dx^2}&=24x^2-6\\ y'(1)&=8-6=2\\ y''(1)&=24-6=18. \end{aligned}

Answer. First derivative $8x^3-6x$ ; second derivative $24x^2-6$ . Their values at $1$ are $2$ and $18$ .

Check. The gradient changes from $y'(0.999)\approx1.98202$ to $y'(1.001)\approx2.01802$ . Its change divided by $0.002$ is approximately $18.0000$ , matching the second derivative.

Change only the exponent in the middle term from $2$ to $-2$ . Track its signs through both derivatives.

Worked example 6: differentiating a reciprocal twice

Question. For $y=2x^4-3x^{-2}+7$ , $x\ne0$ , find the first and second derivatives and their values at $x=1$ .

Plan. Apply the power rule separately at each differentiation.

\begin{aligned} \frac{dy}{dx}&=8x^3+6x^{-3}\\ \frac{d^2y}{dx^2}&=24x^2-18x^{-4}\\ y'(1)&=8+6=14\\ y''(1)&=24-18=6. \end{aligned}

Answer. $dy/dx=8x^3+6/x^3$ and $d^2y/dx^2=24x^2-18/x^4$ , for $x\ne0$ . At $1$ , they are $14$ and $6$ .

Check. A centred change in the first derivative from $0.999$ to $1.001$ , divided by $0.002$ , is approximately $5.99995$ , close to $6$ .

The opening function $x^{3/2}$ now needs only one line: its gradient is $\frac32\sqrt{x}$ for $x>0$ . The limiting meaning remains the same even though the calculation is quicker.

Watch out

For $y=x^4$ , the second derivative is $12x^2$ , not $(4x^3)^2$ . Differentiate the first derivative; do not square it.

Keep $x$ until both differentiation steps are complete.

Quick Recap

For a rational exponent, multiply by $n$ and reduce the exponent by $1$ .
Rewrite roots and reciprocal terms as powers before differentiating.
Keep constant multipliers; the derivative of a constant is zero.
Differentiate again for a second derivative, then evaluate at the required input.
Preserve the original domain and check where the derivative is finite.

For $y=x^{3/2}$ , the tangent gradient is $\frac32\sqrt{x}$ on $x>0$ .

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1.7 - Differentiating powers

The power rule

Sums and multiples

Repeated differentiation

Quick Recap

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