Pure 7.2 - Derivatives of standard functions | Edexcel A-Level Mathematics

Build a small, reliable toolkit for differentiating powers, exponentials, logarithms and trigonometric functions. Learn to rewrite an expression before you differentiate, keep its real domain in view, and recognise which formulae you need to remember.

The core differentiation rules

How can a graph keep exactly the same shape when you differentiate it? The function $e^x$ does that. Before reaching this unusual example, start with a familiar graph: $x^3$ has gradient function $3x^2$ . Its power has decreased by one.

You already know

First principles gives $\frac{\mathrm{d}}{\mathrm{d}x}(x^2)=2x$ and $\frac{\mathrm{d}}{\mathrm{d}x}(x^3)=3x^2$ . Index laws give $\sqrt{x}=x^{1/2}$ and $1/x^2=x^{-2}$ .

The same power rule works for rational powers on intervals where the real function is differentiable. Constants contribute zero; sums can be differentiated term by term.

Power rule and linearity

For rational $n$ , on an interval where the real power is differentiable,

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

For differentiable $f,g$ and constant $c$ ,

(cf)'=cf',\qquad (f\pm g)'=f'\pm g',\qquad \frac{\mathrm{d}}{\mathrm{d}x}(c)=0

Not in the formula booklet: learn it.

For $x>0$ , every rational power is straightforward. Even-root functions need a non-negative input to be real, and their derivatives may fail at zero. Negative powers exclude zero; odd roots can allow negative inputs. Thus $x^{1/3}$ exists at zero but has no finite derivative there, whereas $x^{4/3}$ has derivative zero there. Check the expression, not just its appearance.

Worked example 1: fractional and negative powers

Question. Differentiate $y=5x^4-3x^{1/2}+8x^{-2}$ , for $x>0$ .

Plan. Apply the power rule to each term with its coefficient and sign attached.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=5(4)x^3-3\left(\frac12\right)x^{-1/2}+8(-2)x^{-3} && \text{lower each power by one}\\ &=20x^3-\frac32x^{-1/2}-16x^{-3} && \text{collect coefficients} \end{aligned}

Why this step? A negative exponent decreases too: $-2-1=-3$ , not $-1$ .

Answer. $20x^3-\frac32x^{-1/2}-16x^{-3}$ .

Check. At $x=1$ , the gradient is $20-3/2-16=5/2$ ; all terms are defined on the stated domain.

Change only the middle power from $1/2$ to $3/2$ . This changes both its multiplier and the remaining power.

Worked example 2: a larger fractional power

Question. Differentiate $y=5x^4-3x^{3/2}+8x^{-2}$ , for $x>0$ .

Plan. Keep the two unchanged derivative terms and recompute the middle one.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=20x^3-3\left(\frac32\right)x^{1/2}-16x^{-3} && \text{power rule}\\ &=20x^3-\frac92x^{1/2}-16x^{-3} && \text{simplify} \end{aligned}

Why this step? The derivative of $x^{3/2}$ is $\frac32x^{1/2}$ , not $\frac32x^{3/2}$ .

Answer. $20x^3-\frac92\sqrt{x}-16x^{-3}$ .

Check. At $x=1$ , its value is $-1/2$ , exactly $3$ below the first example's derivative there.

Do not turn the special case of a constant into a negative power: the derivative of $8$ is zero.

Watch out

Wrong: $(x^{-2})'=-2x^{-1}$ . The power is reduced by $1$ , so the correct derivative is $-2x^{-3}$ . Bring the old power down, then subtract one.

Change the coefficient of the same fractional term once more.

Prepare algebraic expressions first

A product can hide a simple polynomial. For example,

(2x+5)(x-1)=2x^2+3x-5

so its derivative is $4x+3$ . Expanding first avoids learning a new rule for this calculation. For a fraction with a single power downstairs, divide each numerator term by that power instead.

Sum of powers

A sum of powers of $x$ has terms of the form $cx^n$ , with constant $c,n$ . For example, $\frac14+\frac34x^{-1}-\frac54x^{-2}$ is such a sum. The expression $1/(x+1)$ is not a single power of $x$ .

Your first decision is algebraic: make the power rule visible without changing the original domain.

Worked example 3: dividing by a square

Question. Differentiate $y=(x^2+3x-5)/(4x^2)$ , for $x>0$ .

Plan. Split the fraction into powers, then differentiate term by term.

\begin{aligned} y&=\frac14+\frac34x^{-1}-\frac54x^{-2} && \text{subtract exponents}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &=-\frac34x^{-2}+\frac52x^{-3} && \text{power rule} \end{aligned}

Why this step? The denominator is one power, so each term can be divided separately.

Answer. $-\frac{3}{4x^2}+\frac{5}{2x^3}$ .

Check. At $x=1$ , the derivative is $-3/4+5/2=7/4$ ; the constant $1/4$ has disappeared.

Change only the denominator's power from $2$ to $1/2$ . This is the same rewriting method, now with fractional exponents.

Worked example 4: dividing by a square root

Question. Differentiate $y=(x^2+3x-5)/(4\sqrt{x})$ , for $x>0$ .

Plan. Write the denominator as $4x^{1/2}$ and subtract $1/2$ from each numerator exponent.

\begin{aligned} y&=\frac14x^{3/2}+\frac34x^{1/2}-\frac54x^{-1/2} && \text{index laws}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &=\frac38x^{1/2}+\frac38x^{-1/2}+\frac58x^{-3/2} && \text{power rule} \end{aligned}

Why this step? The negative coefficient and negative exponent in the last term give a positive derivative term.

Answer. $\frac38x^{1/2}+\frac38x^{-1/2}+\frac58x^{-3/2}$ .

Check. At $x=1$ , the derivative is $11/8$ . Zero is excluded because the original denominator vanishes there.

Combining the earlier product and fraction is still just term-by-term differentiation. For example, $(3x-1)(x+4)+(x^2-2x+5)/(2x^2)$ rewrites as $3x^2+11x-7/2-x^{-1}+(5/2)x^{-2}$ , whose derivative is $6x+11+x^{-2}-5x^{-3}$ .

Watch out

Wrong: $x^2/\sqrt{x}=x^{5/2}$ . Division subtracts exponents, so it is $x^{3/2}$ . Also, cancelling factors never restores an input excluded by the original expression.

Change one numerator coefficient in the square-root fraction.

Exponential and logarithmic rules

The opening surprise has a precise form: the height of $e^x$ equals its gradient at every real input. For other positive bases, a constant factor $\ln a$ appears. This is why the base $e$ is so useful in growth models.

Exponential and logarithmic derivatives

For constant $k$ and positive constant $a$ ,

\frac{\mathrm{d}}{\mathrm{d}x}(e^{kx})=ke^{kx},\qquad \frac{\mathrm{d}}{\mathrm{d}x}(a^{kx})=k(\ln a)a^{kx}

For $x>0$ ,

\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac1x

Not in the formula booklet: learn it.

The booklet supplies $e^{x\ln a}=a^x$ , which explains the extra factor: $a^{kx}=e^{(k\ln a)x}$ . It does not print the derivative rules above. If $a=1$ , the function is constant and $\ln1=0$ , so the same formula gives zero.

Worked example 5: three rule families

Question. Differentiate $y=4e^{-3x}-2^{5x}+3\ln x$ , for $x>0$ .

Plan. Differentiate each term, including the exponent multiplier and the logarithm of the base.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=4(-3)e^{-3x}-5(\ln2)2^{5x}+\frac3x && \text{apply the three rules}\\ &=-12e^{-3x}-5(\ln2)2^{5x}+\frac3x && \text{simplify} \end{aligned}

Why this step? The variable is in an exponent in $2^{5x}$ ; the power rule for $x^n$ does not apply.

Answer. $-12e^{-3x}-5(\ln2)2^{5x}+3/x$ .

Check. $2^{5x}=e^{(5\ln2)x}$ gives exactly the same middle derivative. The logarithm requires $x>0$ .

Change only the base $2$ to $3$ . Both the exponential and its logarithmic multiplier change.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 7.2 - Derivatives of standard functions

The core differentiation rules

Prepare algebraic expressions first

Exponential and logarithmic rules

Continue this lesson.

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