P3.4 - Differentiation

Learning objectives

P3.4.1Differentiating exp, log and trig functionsDifferentiation of ekx, ln kx, sin kx, cos kx, tan kx and their sums and differences.P3.4.2DifferentiationDifferentiation using the product Differentiation of cosec x, cot x and sec x are required. rule, the quotient rule and the chain Skill will be expected in the differentiation of functions rule. generated from standard functions using products, quotients and composition, such as e3x 2x4 sin x,, cos x2 and tan2 2x. x.P3.4.3Reciprocal derivative relationshipUse dy/dx = 1/(dx/dy), including finding dy/dx when x is given as a function of y.P3.4.4exponentialUnderstand and use exponential Students should be familiar with terms such as ‘initial’, growth and decay. ‘meaning when’ t = 0.; Students may need to explore the behaviour for large values of t or to consider whether the range of values predicted is appropriate.; Consideration of a second improved model may be required. d Knowledge and use of the result (ax) = ax ln a is dx expected.

Differentiate a sum or difference term by term. For each exponential, logarithmic or trigonometric term, multiply by the derivative of its linear inner expression.

$f(x)$	$f'(x)$
$e^{kx}$	$ke^{kx}$
$\ln(kx)$	$1/x$ on its domain
$\sin(kx)$	$k\cos(kx)$
$\cos(kx)$	$-k\sin(kx)$
$\tan(kx)$	$k\sec^2(kx)$

For $y=3e^{-2x}+4\ln(5x)-2\cos(3x),\qquad x>0,$ differentiate each term: $\frac{dy}{dx}=-6e^{-2x}+\frac4x+6\sin(3x).$ The domain condition comes from the logarithm.

These trigonometric derivatives assume angles are measured in radians. A derivative may then be evaluated at a point to obtain a tangent gradient or set equal to zero to locate stationary candidates.

For $\ln(kx)$ , the chain factor cancels algebraically to $1/x$ , but the original condition $kx>0$ remains. Do not omit the inner factor $k$ from $e^{kx}$ , sine, cosine or tangent.

The structure of the expression selects the rule: multiplication needs the product rule, division needs the quotient rule, and a function inside another function needs the chain rule. More than one rule may be nested in the same derivative.

Structure	Derivative
$uv$	$u'v+uv'$
$u/v$	$(vu'-uv')/v^2$
$f(g(x))$	$f'(g(x))g'(x)$

$\frac{d}{dx}(\sec x)=\sec x\tan x,\qquad \frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\qquad \frac{d}{dx}(\cot x)=-\operatorname{cosec}^2x$

Examples expose the chosen structure: $\frac{d}{dx}(2x^4\sin x)=8x^3\sin x+2x^4\cos x,$ $\frac{d}{dx}\left(\frac{e^{3x}}x\right)=\frac{e^{3x}(3x-1)}{x^2},$ $\frac{d}{dx}\big(\tan^2(2x)\big)=4\tan(2x)\sec^2(2x).$

For a nested expression, identify the outermost operation first, write its rule without simplifying, and then differentiate each inner part. Factor common terms afterwards when solving a stationary-point equation.

The quotient numerator order is denominator times derivative of numerator minus numerator times derivative of denominator. Do not differentiate a product by multiplying derivatives, and retain every original domain restriction.

If a curve is given as $x=f(y)$ , differentiate with respect to $y$ first and then take the reciprocal. This works on a local inverse branch wherever $dx/dy\ne0$ .

$\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}$

Use this order: find dx/dy; invert it; simplify in terms of y; then, only if required, use the original relation and the stated interval to rewrite the result in terms of x. The interval controls any square-root sign.

Suppose $x=\sin(3y)$ on $-\frac\pi6<y<\frac\pi6$ . Then $\frac{dx}{dy}=3\cos(3y),\qquad \frac{dy}{dx}=\frac1{3\cos(3y)}.$ On this branch cosine is positive and $\cos(3y)=\sqrt{1-x^2}$ , so $\frac{dy}{dx}=\frac1{3\sqrt{1-x^2}}.$

After evaluating $dy/dx$ at a point, the normal gradient is its negative reciprocal. Keep the point coordinates with the gradient when forming a tangent or normal equation.

The symbols $dy/dx$ and $dx/dy$ are reciprocals only where a differentiable local inverse exists. If $dx/dy=0$ , this calculation does not produce a finite $dy/dx$ ; do not divide by zero or discard the branch condition.

An exponential model describes change at a rate proportional to its current displacement from a limiting level. Read the initial value at t=0, the sign of the rate, and the long-term behaviour before trusting a numerical prediction.

$\frac{d}{dx}(a^x)=a^x\ln a\quad(a>0),\qquad \frac{d}{dt}(Ae^{kt}+C)=kAe^{kt}=k(y-C)$

For $A>0$ :

Model feature	Interpretation
$y(0)=A+C$	initial value
$k>0$	growth away from $C$
$k<0$	decay towards $C$ as $t\to\infty$
$y=Ab^t$	growth if $b>1$ ; decay if $0<b<1$

For a temperature model $T=20+80e^{-0.3t},\qquad t\ge0,$ the initial temperature is $100$ , the limiting temperature is $20$ , and $\frac{dT}{dt}=-24e^{-0.3t}=-0.3(T-20).$ Thus the temperature falls quickly at first and the magnitude of its rate tends to zero.

A model's algebra may be correct but its prediction may be inappropriate outside the observed time range: unlimited growth may exceed a physical capacity, while a decay limit may be unrealistic. Compare an improved model by its fit in the relevant range and by whether its initial value, limit and allowed outputs make sense.

Time models normally use $t\ge0$ . Do not call every decreasing exponential a decay to zero: an outside constant $C$ changes the limit, and the sign of $A$ also affects whether the curve approaches $C$ from above or below.

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—P3.5 - IntegrationRead

P3.4 - Differentiation

Learning objectives

Differentiate exponential, logarithmic and trigonometric terms

Choose and nest the product, quotient and chain rules

Differentiate when x is given as a function of y

Interpret and test exponential growth or decay models

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