G2 - Standard derivatives | AQA A-Level Mathematics

Differentiate rational powers, exponentials, logarithms and trigonometric functions with linear arguments. Combine the rules in sums, keep the real domain in view and use numerical gradients as checks on your algebra.

Differentiating rational powers

Can you find the slope of $4\sqrt x-3/x^2+2\sqrt[3]x$ without expanding a difference quotient full of roots? At $x=1$ , its three terms have gradients $2$ , $6$ and $2/3$ , giving total slope $26/3$ . Writing roots and reciprocals as powers makes one rule do all three jobs.

You already know

$\sqrt x=x^{1/2}$ , $\sqrt[3]x=x^{1/3}$ and $1/x^2=x^{-2}$ . For positive integer powers, first principles gives $d(x^n)/dx=nx^{n-1}$ . These index and derivative results are not supplied in the formula booklet.

The power rule extends to rational exponents wherever the real function is differentiable. Multiply by the original exponent, then subtract one from the exponent.

Rational-power rule

For rational $n$ and constant $A$ , where the derivative exists,

\frac{d}{dx}(Ax^n)=Anx^{n-1}

Constants have derivative zero; for $n=0$ , differentiate the constant directly rather than interpreting $0x^{-1}$ at zero.
Not in the formula booklet: learn it.

For example, $d(x^{-2})/dx=-2x^{-3}$ . Express a rational exponent in lowest terms before judging its real domain. An even root denominator requires nonnegative input; a negative exponent excludes zero. Odd roots allow negative inputs, but the derivative can still fail at zero: $d(\sqrt[3]x)/dx=1/(3x^{2/3})$ for $x\ne0$ , with no finite derivative at zero. Similarly, $\sqrt x$ exists at zero but its derivative does not. Conversely, $x^{4/3}$ has derivative zero at zero, as its difference quotient there is $h^{1/3}\to0$ .

Worked example 1: roots and a reciprocal

Question. Differentiate $y=4\sqrt x-3/x^2+2\sqrt[3]x$ , for $x>0$ .

Plan. Rewrite all three terms as powers, apply the rational-power rule term by term and preserve the domain.

\begin{aligned} y&=4x^{1/2}-3x^{-2}+2x^{1/3} && \text{index notation}\\ \frac{dy}{dx}&=4\left(\frac12\right)x^{-1/2}-3(-2)x^{-3}+2\left(\frac13\right)x^{-2/3} && \text{power rule}\\ &=2x^{-1/2}+6x^{-3}+\frac23x^{-2/3} && \text{simplify coefficients} \end{aligned}

Why this step? Rewriting the reciprocal exposes exponent $-2$ ; subtracting one makes it $-3$ , rather than $-1$ .

Answer. $dy/dx=2/\sqrt x+6/x^3+2/(3x^{2/3})$ , for $x>0$ .

Check. At $x=1$ , the three contributions sum to $2+6+2/3=26/3$ , answering the opening slope question.

Change only the reciprocal's exponent, from $2$ to $3$ . Its derivative changes in both coefficient and exponent; the root terms stay as they were.

Worked example 2: a different reciprocal power

Question. Differentiate $y=4\sqrt x-3/x^3+2\sqrt[3]x$ , for $x>0$ .

Plan. Rewrite the reciprocal as $-3x^{-3}$ , then differentiate each power.

\begin{aligned} y&=4x^{1/2}-3x^{-3}+2x^{1/3} && \text{rewrite}\\ \frac{dy}{dx}&=2x^{-1/2}+9x^{-4}+\frac23x^{-2/3} && \text{multiply, then subtract one}\\ y'(1)&=2+9+\frac23=\frac{35}{3} && \text{evaluate after differentiating} \end{aligned}

Why this step? The minus sign of the coefficient and the negative exponent cancel when multiplying; the derivative of the negative reciprocal is positive here.

Answer. $dy/dx=2x^{-1/2}+9x^{-4}+(2/3)x^{-2/3}$ , for $x>0$ .

Check. Only the middle slope at $1$ changed, from $6$ to $9$ , so the total increased by $3$ .

You can often simplify a fraction before differentiating: $(2x^2+3\sqrt x)/x=2x+3x^{-1/2}$ for $x>0$ . Its derivative is $2-(3/2)x^{-3/2}$ ; the original restriction still applies. These are power-rule consequences, not separately supplied formulae.

Watch out

Wrong: $d(x^{-3})/dx=-3x^{-2}$ . Subtracting one from $-3$ gives $-4$ . Right: $d(x^{-3})/dx=-3x^{-4}$ .

Change the reciprocal exponent once more, keeping the two root terms fixed.

Exponential and logarithmic derivatives

In $x^3$ , the variable is the base; in $3^x$ , it is the exponent. These are different functions and need different rules. The special exponential $e^x$ is its own gradient function: at zero its height and slope are both $1$ , and at $1$ both are $e$ . A curve can carry a copy of itself in its tangent slopes.

Exponential and logarithmic rules

For constants $k,b$ ,

\frac{d}{dx}e^{kx+b}=ke^{kx+b} \qquad \frac{d}{dx}a^{kx+b}=k\ln(a)\,a^{kx+b} \qquad \frac{d}{dx}\ln x=\frac1x

Use $a>0$ , and $x>0$ for $\ln x$ . The case $a=1$ is the constant function $1$ , with derivative zero.
Not in the formula booklet: learn it.

For $a=e$ , the factor $\ln e=1$ recovers the first rule. For $0<a<1$ , $\ln a<0$ , explaining why $a^x$ decreases. The factor $k$ measures the constant change of the linear exponent; adding the fixed shift $b$ does not change it. These listed linear forms can be used now; more general composite functions need a later method.

Worked example 3: an exponential and a logarithm

Question. Find $f'(x)$ for $f(x)=5e^{2x}+\ln x$ , where $x>0$ .

Plan. Apply the exponential rule with $k=2$ and add the logarithmic derivative.

\begin{aligned} f'(x)&=5(2e^{2x})+\frac1x && \text{differentiate both terms}\\ &=10e^{2x}+\frac1x && \text{simplify}\\ f'(1)&=10e^2+1 && \text{substitute afterwards} \end{aligned}

Why this step? The multiplier $2$ belongs to the exponent's rate of change; it remains even though the exponential itself stays in the answer.

Answer. $f'(x)=10e^{2x}+1/x$ , on $x>0$ .

Check. Both contributions are positive there, so the sum is increasing; at $1$ , the slope is approximately $74.8906$ .

Change only the exponential base from $e$ to $3$ . The factor $\ln3$ now matters.

Worked example 4: a positive base other than e

Question. Find $g'(x)$ for $g(x)=5\times3^{2x}+\ln x$ , where $x>0$ .

Plan. Use the same exponent factor $2$ , include $\ln3$ , and differentiate the logarithm as before.

\begin{aligned} g'(x)&=5[2\ln(3)\,3^{2x}]+\frac1x && \text{other-base rule}\\ &=10\ln(3)\,3^{2x}+\frac1x && \text{simplify}\\ g'(1)&=90\ln3+1 && 3^2=9 \end{aligned}

Why this step? The exponential's base is no longer $e$ , so omitting $\ln3$ would predict the wrong gradients everywhere.

Answer. $g'(x)=10\ln(3)\,3^{2x}+1/x$ , on $x>0$ .

Check. $3^{2x}=9^x$ gives derivative $5\ln(9)\,9^x=10\ln(3)\,3^{2x}$ , the same exponential term. At $1$ , the full slope is approximately $99.8751$ .

For a mixture of bases, keep each rule with its own term: $5e^{2x}-7\times3^x+\ln x$ differentiates to $10e^{2x}-7\ln(3)\,3^x+1/x$ . The logarithm still excludes nonpositive inputs; its reciprocal derivative cannot create a logarithm value at a negative input.

Watch out

Wrong: $d(3^{2x})/dx=2\times3^{2x}$ . This forgets the base factor. Right: $d(3^{2x})/dx=2\ln(3)\,3^{2x}$ . Also, $e^2$ has derivative zero because it is constant in $x$ .

Change the base once more, to a number between zero and one, and inspect the sign of its contribution.

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G2 - Standard derivatives

Differentiating rational powers

Exponential and logarithmic derivatives

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