1.07j, 1.07l - Differentiating exponential and logarithmic functions | OCR A-Level Mathematics A

Find gradients of natural and other positive-base exponentials with linear exponents (1.07j), and of the natural logarithm on its real domain (1.07l). You will combine these rules in sums and distinguish an instantaneous rate from a change over a whole time interval.

The natural exponential and its rate factor

If an amount decreases at one quarter of its current amount per hour, does it lose exactly a quarter during the next hour? For $C(t)=80e^{-t/4}$ , the answer is no. Differentiation will explain the difference between a rate now and a loss over an hour.

First consider $y=e^x$ . At $x=0$ , its value and gradient are both 1. At $x=1$ , both are $e$ . This remarkable equality holds at every point, not just at those two inputs.

You already know

$e^x$ is the natural exponential, and $\ln x$ its inverse. Distinguish $x^n$ , with a constant exponent, from $a^x$ , with a constant base. Differentiate sums and constant multiples term by term.

The difference quotient makes the special gradient property visible:

\begin{aligned} \frac{e^{x+h}-e^x}{h}&=e^x\frac{e^h-1}{h} &&\text{factor out }e^x\\ \lim_{h\to0}\frac{e^{x+h}-e^x}{h}&=e^x &&\text{since }\frac{e^h-1}{h}\to1. \end{aligned}

For $e^{kx}$ , set $u=kh$ in the same limiting argument: the extra factor is $k$ , not the whole exponent $kx$ . The first-principles definition is in the formula sheet; the exponential limit and derivative below are not.

Natural exponential derivative

For a real constant $k$ :

\frac{\mathrm d}{\mathrm dx}e^x=e^x,\qquad \frac{\mathrm d}{\mathrm dx}e^{kx}=ke^{kx}.

The domain is all real $x$ . For $k=0$ , the function is constant and the formula gives zero.

Not in the formula sheet: learn it.

Think of the factor $k$ as the speed at which the input changes. General composite functions come later; here the input is linear.

Worked example 1: a negative input rate

Question. Differentiate $y=5e^{-2x}$ and find $y'(0)$ .

Plan. Retain the coefficient 5 and multiply by the exponent's constant rate $-2$ .

\begin{aligned} y'&=5(-2)e^{-2x} &&\text{input rate is }-2\\ &=-10e^{-2x} &&\text{simplify}\\ y'(0)&=-10 &&\text{since }e^0=1 \end{aligned}

Why this step? Lowering the exponent would apply a power rule to the wrong kind of function.

Answer. $y'=-10e^{-2x}$ , with gradient $-10$ at zero.

Check. The function is positive and decreasing, matching the negative derivative.

Change only the exponent's sign. The height remains positive, but its direction reverses.

Worked example 2: a positive input rate

Question. Differentiate $y=5e^{2x}$ and find $y'(0)$ .

Plan. Multiply by the input rate 2, then substitute zero.

\begin{aligned} y'&=5(2)e^{2x} &&\text{linear input rate}\\ &=10e^{2x} &&\text{collect the constants}\\ y'(0)&=10 &&\text{evaluate the derivative} \end{aligned}

Why this step? Changing $-2x$ to $2x$ changes the derivative's sign as well as the exponent.

Answer. $y'=10e^{2x}$ , with gradient 10 at zero.

Check. The gradient is positive for every real input, consistent with exponential growth.

Keep the power rule and exponential rule distinct.

Watch out

Wrong: $\frac{\mathrm d}{\mathrm dx}e^{2x}=2x e^{2x-1}$ . The exponent is variable, so this is not the power rule. Multiply by the constant input rate 2 and retain the exponential: $2e^{2x}$ .

A fractional input rate works in the same way: for example, $3e^{x/2}-7$ has derivative $\tfrac32e^{x/2}$ . The constant disappears.

Other bases introduce a logarithm factor

For another positive base, use the natural exponential as a bridge: $a^{kx}=e^{(k\ln a)x}$ . Its input rate is $k\ln a$ .

Derivative for a positive base

For $a>0$ and real constant $k$ :

\frac{\mathrm d}{\mathrm dx}a^{kx}=k\ln(a)\,a^{kx}.

This includes $a=1$ , where $\ln1=0$ . A negative base does not give a real exponential for all real inputs.

Not in the formula sheet: learn it.

The logarithm belongs to the base. It is a fixed number, even though the input $x$ varies.

Worked example 3: a base above one

Question. Differentiate $y=2^{3x}$ and find its exact gradient at zero.

Plan. Rewrite with base $e$ , then use the input rate.

\begin{aligned} y&=e^{(3\ln2)x} &&\text{change the exponential base}\\ y'&=3\ln2\,2^{3x} &&\text{differentiate}\\ y'(0)&=3\ln2 &&\text{evaluate} \end{aligned}

Why this step? Rewriting the exponent shows exactly where the factor $\ln2$ comes from.

Answer. $y'=3\ln2\,2^{3x}$ , with gradient $3\ln2$ at zero.

Check. $3\ln2>0$ , consistent with an increasing exponential.

Change only the base to its reciprocal. The logarithm factor should reverse sign.

Worked example 4: a base below one

Question. Differentiate $y=(\tfrac12)^{3x}$ and find its exact gradient at zero.

Plan. Use the same rule and simplify the logarithm of the base.

\begin{aligned} y'&=3\ln(\tfrac12)(\tfrac12)^{3x} &&\text{positive-base rule}\\ &=-3\ln2\,(\tfrac12)^{3x} &&\text{logarithm of a reciprocal}\\ y'(0)&=-3\ln2 &&\text{evaluate} \end{aligned}

Why this step? A positive exponential value multiplied by a negative logarithm gives a negative gradient everywhere.

Answer. $y'=-3\ln2\,(\tfrac12)^{3x}$ , with gradient $-3\ln2$ at zero.

Check. $(\tfrac12)^{3x}=2^{-3x}$ gives the same derivative directly.

The two examples have the same base-logarithm magnitude but opposite signs.

Watch out

The extra factor is $\ln a$ , not $\ln x$ . The base is fixed, so its logarithm is a constant. Missing that factor gives the wrong gradient even when the exponential part is correct.

In general, it is the sign of $k\ln a$ that controls direction. For example, $(1/4)^x$ has derivative $-\ln4\,(1/4)^x$ . Do not confuse $2^x$ with $x^2$ : their derivatives are $\ln2\,2^x$ and $2x$ .

The natural logarithm and its real domain

The logarithm's reciprocal gradient comes from its inverse relationship with the exponential. Reflection in $y=x$ exchanges horizontal and vertical changes, so corresponding tangent gradients are reciprocals.

Exponential and logarithm tangent gradients at corresponding inverse points

The exponential has gradient $x$ at the point $(\ln x,x)$ ; the logarithm therefore has gradient $1/x$ at $(x,\ln x)$ .

Natural logarithm derivative

\frac{\mathrm d}{\mathrm dx}\ln x=\frac1x,\qquad x>0.

For positive constant $c$ , $\ln(cx)=\ln c+\ln x$ on $x>0$ , so its derivative is also $1/x$ .

Not in the formula sheet: learn it.

The expression $1/x$ exists for negative $x$ , but that does not extend the real domain of $\ln x$ . On its actual domain, the logarithm increases while its gradient becomes smaller.

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1.07j, 1.07l - Differentiating exponential and logarithmic functions

The natural exponential and its rate factor

Other bases introduce a logarithm factor

The natural logarithm and its real domain

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