P1 4.2, P1 4.1 - Differentiating powers | Edexcel A-Level International Mathematics

Find gradient functions efficiently by differentiating powers and sums. Learn how to rewrite products, quotients and roots first, then differentiate once more to obtain a second derivative.

Power rule and sums

How could you find the gradient function of $y=(x+2)(x-3)/\sqrt{x}$ without a long chord calculation? The key is to rewrite it as powers whose derivatives follow one simple rule.

For $y=x^2$ , chord gradients gave $\mathrm{d}y/\mathrm{d}x=2x$ . For $y=x^3$ , the same calculation at $x=a$ gives

\frac{(a+h)^3-a^3}{h}=3a^2+3ah+h^2,

which approaches $3a^2$ as $h$ approaches zero. The exponent becomes a multiplier, then decreases by one.

Power rule for differentiation

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

This applies wherever the power is real and differentiable. Multiply by any constant coefficient, and differentiate sums and differences term by term:

\frac{\mathrm{d}}{\mathrm{d}x}(kx^n)=knx^{n-1}.

A constant has derivative $0$ , and $\mathrm{d}(kx)/\mathrm{d}x=k$ .

Not in the formula booklet: learn it.

A constant shifts a graph vertically without changing its tangent gradients. Likewise, multiplying all heights by $k$ multiplies all vertical changes, so it multiplies the derivative by $k$ . For a sum, the changes add, so the derivatives add.

Worked example 1: differentiate a polynomial

Question. Find $\mathrm{d}y/\mathrm{d}x$ when $y=6x^3-5x^2+4x-9$ .

Plan. Apply the power rule to each term and remove the constant term.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=6(3)x^{3-1}-5(2)x^{2-1}+4\\ &=18x^2-10x+4 \end{aligned}

Answer. $\mathrm{d}y/\mathrm{d}x=18x^2-10x+4$ .

Check. At $x=0$ , the terms in $x^3$ and $x^2$ have horizontal tangents, leaving the linear term's gradient $4$ . The result gives $4$ .

Change only the first exponent from $3$ to $3/2$ . The same two operations apply to a fractional index.

Worked example 2: a fractional power

Question. Find $\mathrm{d}y/\mathrm{d}x$ when $y=6x^{3/2}-5x^2+4x-9$ , for $x>0$ .

Plan. Multiply by each exponent, then subtract one from it.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=6\left(\frac32\right)x^{3/2-1}-10x+4\\ &=9x^{1/2}-10x+4 \end{aligned}

Answer. $\mathrm{d}y/\mathrm{d}x=9\sqrt{x}-10x+4$ .

Check. At $x=4$ , the result gives $-18$ . The symmetric nearby quotient $[y(4.001)-y(3.999)]/0.002$ is approximately $-18.00000005$ , agreeing to the displayed precision.

Negative indices work too: $\mathrm{d}(x^{-2})/\mathrm{d}x=-2x^{-3}$ for $x\ne0$ . Keep the sign of the old exponent when forming the new coefficient.

Watch out

$\mathrm{d}(6x^{3/2})/\mathrm{d}x=9x^{1/2}$ , not $9x^{3/2}$ . Multiplying by the exponent and reducing it are both necessary. Also, the derivative of $-9$ is $0$ , not $-9$ .

Use the same term-by-term method with a new first coefficient.

Rewriting products quotients and roots before differentiation

A product or quotient may hide a sum of powers; expose those powers before differentiating.

For example, $\sqrt{x}=x^{1/2}$ and $1/x^2=x^{-2}$ . A single-power denominator can be divided into every term separately:

\frac{x^2+5x-3}{3\sqrt{x}} =\frac13x^{3/2}+\frac53x^{1/2}-x^{-1/2} \qquad(x>0).

The quotient law subtracts the denominator's exponent from each numerator exponent. The coefficient $1/3$ also applies to every term.

You already know

Expand brackets before collecting powers. For $x>0$ , $x^a/x^b=x^{a-b}$ , so dividing by $\sqrt{x}$ subtracts $1/2$ from each exponent.

Start with a product that expands to a quadratic.

Worked example 3: expand a product

Question. Differentiate $y=(x+2)(x-3)$ .

Plan. Expand and collect terms, then use the power rule.

\begin{aligned} y&=x^2-3x+2x-6=x^2-x-6\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=2x-1 \end{aligned}

Answer. $\mathrm{d}y/\mathrm{d}x=2x-1$ .

Check. The expanded graph is $(x-1/2)^2-25/4$ . At its symmetry line $x=1/2$ , the tangent is horizontal, and $2(1/2)-1=0$ .

Change only the denominator: divide the same product by $\sqrt{x}$ . Expand first as before, then divide each term.

Worked example 4: divide by a root

Question. Differentiate $y=(x+2)(x-3)/\sqrt{x}$ , for $x>0$ .

Plan. Expand the numerator, rewrite the quotient as three powers, then differentiate each.

\begin{aligned} y&=\frac{x^2-x-6}{x^{1/2}}\\ &=x^{3/2}-x^{1/2}-6x^{-1/2}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &=\frac32x^{1/2}-\frac12x^{-1/2}+3x^{-3/2} \end{aligned}

Why rewrite first? The power rule applies to each separate power. Differentiating the numerator and denominator separately and dividing their derivatives is not a differentiation rule.

Answer. $\mathrm{d}y/\mathrm{d}x=\frac32\sqrt{x}-\frac1{2\sqrt{x}}+\frac3{x^{3/2}}$ .

Check. At $x=4$ , this gives $3-1/4+3/8=25/8$ . A symmetric quotient from the original expression at $4\pm0.001$ is approximately $3.125000005$ , agreeing with $25/8$ .

The final positive term comes from multiplying the two negative factors $-6$ and $-1/2$ .

Watch out

$\frac{x^2-x-6}{\sqrt{x}}=x^{3/2}-x^{1/2}-6x^{-1/2}$ , not $x^{3/2}-x-6$ . Divide every numerator term. The denominator also excludes $x=0$ , so the original domain matters.

Now change the numerator's two brackets while keeping the same rewriting strategy.

Second derivatives

You can differentiate a gradient function to find how that gradient changes with the input.

For $y=x^3$ , the first derivative is $3x^2$ . Differentiating $3x^2$ gives $6x$ , called the second derivative. You are applying the same operation twice, to two successive functions.

Second derivative

For $y=f(x)$ , the second derivative is the derivative of $f'(x)$ . Write it as

f''(x)\quad\text{or}\quad\frac{\mathrm{d}^2y}{\mathrm{d}x^2}.

It measures the rate of change of the first derivative with respect to $x$ . It is not the square of the first derivative.

Keep a separate line for each derivative so that the second calculation uses the first derivative rather than the original function.

Worked example 5: differentiate twice

Question. Given $f(x)=x^3-3x^2+2x+5$ , find $f''(x)$ and $f''(2)$ .

Plan. Differentiate term by term twice, then substitute $x=2$ into the second result.

\begin{aligned} f'(x)&=3x^2-6x+2\\ f''(x)&=6x-6\\ f''(2)&=12-6=6 \end{aligned}

Answer. $f''(x)=6x-6$ and $f''(2)=6$ .

Check. The first derivative is $g(x)=3x^2-6x+2$ . Its chord gradient from $2$ to $2+h$ is $6+3h$ , which approaches $6$ .

Change only the original leading power from $x^3$ to $x^4$ . Track that change through both derivatives.

Worked example 6: a changed leading power

Question. Given $f(x)=x^4-3x^2+2x+5$ , find $f''(x)$ and $f''(2)$ .

Plan. Find the first and second derivative functions before substituting.

\begin{aligned} f'(x)&=4x^3-6x+2\\ f''(x)&=12x^2-6\\ f''(2)&=12(4)-6=42 \end{aligned}

Answer. $f''(x)=12x^2-6$ and $f''(2)=42$ .

Check. The first derivative's symmetric chord gradient at $2$ , using inputs $2\pm0.001$ , is $42.000004$ , close to $42$ . Squaring $f'(2)=22$ would instead give $484$ , a different calculation.

If the first derivative has units of metres per second and the input is time in seconds, the second derivative has units of metres per second squared. Each differentiation divides the units by the input's units again.

Watch out

Substituting $x=2$ into $f'(x)$ gives a number, not a function you can then differentiate to find $f''(2)$ . Differentiate first; substitute afterwards.

The opening curve needed no new product or quotient rule: expansion and index laws exposed powers. A second derivative follows by applying the same power rule to its gradient function.

Quick Recap

Differentiate $kx^n$ by multiplying by $n$ and reducing the exponent by one.
Differentiate sums term by term; a constant differentiates to zero.
Expand products and rewrite roots or quotients as powers before differentiating.
Differentiate the first derivative to obtain the second derivative, then substitute if a value is required.

The opening curve has gradient function $\frac32x^{1/2}-\frac12x^{-1/2}+3x^{-3/2}$ , for $x>0$ .

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P1 4.2, P1 4.1 - Differentiating powers

Power rule and sums

Rewriting products quotients and roots before differentiation

Second derivatives

Quick Recap

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