Mc10 - Differentiating exponential and logarithmic functions

Mc10 - Differentiating exponential and logarithmic functions

Differentiate ekxe^{kx}, akxa^{kx} and ln⁡x\ln x, together with related sums, differences and constant multiples, as required by Mc10. You will keep the input-rate factors and logarithm domains, and interpret a derivative as an instantaneous rate rather than an average over an interval.

The natural exponential and its rate factor

An amount falls from 120 grams to 60 grams in three hours under the model A(t)=120⋅2−t/3A(t)=120\cdot2^{-t/3}. Is its loss rate therefore always 20 grams per hour? The average loss rate is 20; the instantaneous rate keeps changing. We will calculate it.

Start with the natural exponential. For y=exy=e^x, height and tangent gradient agree at every point: at zero both are 1, and at ln⁡2\ln2 both are 2. Equality of a function and its derivative is a special property, not a rule for all exponentials.

You already know

exe^x and ln⁡x\ln x are inverse functions. Exponential laws give ax=exln⁡aa^x=e^{x\ln a} for a>0a>0. A constant multiple stays as a multiplier when you differentiate.

The first-principles argument separates the current height from one fixed limiting factor:

ex+h−exh=exeh−1hfactoriselim⁡h→0ex+h−exh=exuse lim⁡h→0eh−1h=1.\begin{aligned} \frac{e^{x+h}-e^x}{h}&=e^x\frac{e^h-1}{h} &&\text{factorise}\\ \lim_{h\to0}\frac{e^{x+h}-e^x}{h}&=e^x &&\text{use }\lim_{h\to0}\frac{e^h-1}{h}=1. \end{aligned}

For ekxe^{kx}, substituting u=khu=kh into the corresponding limit introduces a factor kk. The first-principles definition is supplied in the formula sheet, but this limiting value and the derivative rule are not.

Differentiating the natural exponential

For real constant kk:

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

Both functions have domain R\mathbb R. When k=0k=0, the derivative is zero because the function is constant.

Not in the formula sheet: learn it.

The extra factor is the constant rate of the linear input, not the input kxkx itself. General composite-function differentiation comes later.

Worked example 1: a fractional negative input rate

Question. Differentiate y=6e−x/3+4y=6e^{-x/3}+4, and find its gradient at zero.

Plan. Multiply the exponential term by −1/3-1/3; differentiate the constant to zero.

y′=6(−13)e−x/3input rate and constant multiple=−2e−x/3simplifyy′(0)=−2evaluate\begin{aligned} y'&=6(-\tfrac13)e^{-x/3} &&\text{input rate and constant multiple}\\ &=-2e^{-x/3} &&\text{simplify}\\ y'(0)&=-2 &&\text{evaluate} \end{aligned}

Why this step? The exponent changes by −1/3-1/3 per unit change in xx, so it contributes that fixed factor.

Answer. y′=−2e−x/3y'=-2e^{-x/3}; gradient −2-2 at zero.

Check. The curve approaches 4 from above as xx increases, agreeing with a negative gradient.

Change only the exponent's sign. The vertical shift stays constant and still contributes zero.

Worked example 2: reversing the input rate

Question. Differentiate y=6ex/3+4y=6e^{x/3}+4, and find its gradient at zero.

Plan. Use input rate 1/31/3, retaining the coefficient 6.

y′=6(13)ex/3differentiate=2ex/3collect the constantsy′(0)=2substitute into the derivative\begin{aligned} y'&=6(\tfrac13)e^{x/3} &&\text{differentiate}\\ &=2e^{x/3} &&\text{collect the constants}\\ y'(0)&=2 &&\text{substitute into the derivative} \end{aligned}

Why this step? A constant vertical shift changes heights, but never tangent gradients.

Answer. y′=2ex/3y'=2e^{x/3}; gradient 2 at zero.

Check. The derivative is positive everywhere, matching the growing exponential above its baseline.

The exponent's rate, rather than its current value, supplies the multiplier.

Watch out

Wrong: differentiating ex/3e^{x/3} gives (x/3)ex/3(x/3)e^{x/3}. The input x/3x/3 has constant derivative 1/31/3, so the correct result is 13ex/3\tfrac13e^{x/3}.

The factor can be any real constant, including a fraction or zero. Keep it exact rather than rounding it before differentiating.

Other bases introduce a logarithm factor

To differentiate another positive-base exponential, convert it into a natural exponential: akx=e(kln⁡a)xa^{kx}=e^{(k\ln a)x}. Now its linear input rate is visible.

Differentiating another positive base

For positive constant aa and real constant kk:

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

This includes a=1a=1, giving zero because ln⁡1=0\ln1=0. Negative bases are excluded from this all-real-input rule.

Not in the formula sheet: learn it.

It is ln⁡a\ln a, not ln⁡x\ln x, that enters the coefficient. Its sign, together with that of kk, determines growth or decay.

Worked example 3: introducing the logarithm of the base

Question. Differentiate y=32xy=3^{2x}, giving its exact gradient at zero.

Plan. Rewrite with base ee, then differentiate the linear exponent.

y=e(2ln⁡3)xexponential identityy′=2ln⁡3 32xinput rate 2ln⁡3y′(0)=2ln⁡3evaluate exactly\begin{aligned} y&=e^{(2\ln3)x} &&\text{exponential identity}\\ y'&=2\ln3\,3^{2x} &&\text{input rate }2\ln3\\ y'(0)&=2\ln3 &&\text{evaluate exactly} \end{aligned}

Why this step? The base contributes a logarithm factor that would be missed by treating this as x3x^3.

Answer. y′=2ln⁡3 32xy'=2\ln3\,3^{2x}; gradient 2ln⁡32\ln3 at zero.

Check. 2ln⁡3>02\ln3>0, agreeing with the function's increasing direction.

Change only the sign of the input coefficient. The logarithm of the base does not change.

Worked example 4: a negative input coefficient

Question. Differentiate y=3−2xy=3^{-2x}, giving its exact gradient at zero.

Plan. Multiply by the new input rate −2ln⁡3-2\ln3.

y′=−2ln⁡3 3−2xpositive-base ruley′(0)=−2ln⁡3evaluate\begin{aligned} y'&=-2\ln3\,3^{-2x} &&\text{positive-base rule}\\ y'(0)&=-2\ln3 &&\text{evaluate} \end{aligned}

Why this step? The negative input coefficient reverses growth even though the base is still greater than one.

Answer. y′=−2ln⁡3 3−2xy'=-2\ln3\,3^{-2x}; gradient −2ln⁡3-2\ln3 at zero.

Check. 3−2x=(1/9)x3^{-2x}=(1/9)^x, whose base lies between zero and one, confirming decay.

Neither the base nor the exponent's coefficient determines the sign alone.

Watch out

A base greater than one can still give a decreasing function when the input coefficient is negative. Use the sign of kln⁡ak\ln a in the derivative of akxa^{kx}; the exponential factor itself is positive.

A base below one and a negative input coefficient reverse the sign twice. For comparison, (1/4)x(1/4)^x has derivative −ln⁡4 (1/4)x-\ln4\,(1/4)^x; changing its input sign would make its gradient positive.

The natural logarithm and its real domain

Reflection of an exponential graph in y=xy=x gives its inverse logarithm graph. Corresponding tangent gradients are reciprocals because horizontal and vertical changes exchange roles.

Inverse exponential and logarithm points have reciprocal tangent gradients

At (ln⁡2,2)(\ln2,2), the exponential has gradient 2; at (2,ln⁡2)(2,\ln2), the logarithm has gradient 1/21/2. The general version gives the standard derivative.

Natural logarithm derivative

ddxln⁡x=1x,x>0.\frac{\mathrm d}{\mathrm dx}\ln x=\frac1x,\qquad x>0.

For c>0c>0, ln⁡(cx)=ln⁡c+ln⁡x\ln(cx)=\ln c+\ln x on x>0x>0, so ddxln⁡(cx)=1/x\frac{\mathrm d}{\mathrm dx}\ln(cx)=1/x too.

Not in the formula sheet: learn it.

Although 1/x1/x exists at negative inputs, ln⁡x\ln x does not have real values there. The derivative inherits the function's domain. It is positive on that domain and gets smaller as xx increases.

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