Mixed Rules and Stationary Points | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Use the formula sheet

Every inside u contributes u′

eᵘ

→ u′eᵘ

always positive

ln u

→ u′ / u

u > 0

sin u

→ u′ cos u

radians

cos u

→ −u′ sin u

keep the minus

tan u

→ u′ sec²u

radians

Write u first when the inside is more than x.

The formula sheet supplies derivative rules. It does not identify the structure of the expression or add the chain-rule factor for you.

Keep nested rules separate

Worked example

A current-paper quotient

y=\frac{\ln(2x+1)}{x+3}

u'=\frac2{2x+1},\qquad v'=1

y'=\frac{\frac{2(x+3)}{2x+1}-\ln(2x+1)}{(x+3)^2}

The quotient rule earns the method mark; the derivative $2/(2x+1)$ is the chain-rule step inside it. (9709/22/M/J/24 Q6(a))

Form the stationary equation

Worked example

Exponential factor followed by a trig equation

y=e^x\cos x

y'=e^x(\cos x-\sin x)

Because $e^x\ne0$ , stationary points satisfy $\cos x-\sin x=0$ , so $\tan x=1$ . Now solve that trig equation over the stated interval.

Examiner note

Keep exact values unless the question asks for a decimal. If an angle must be approximated, solve over the full interval and round only the final answers.

Interpret the derivative sign

A derivative changing from positive to negative means the graph rises and then falls, so the turning point is a local maximum. A change from negative to positive gives a local minimum.

Your turnIndependent check [6 marks]

For $y=xe^{-x}$ , find the stationary point and use the sign of $dy/dx$ to determine its nature.

Show worked answer

y'=e^{-x}(1-x)

Since $e^{-x}>0$ , the derivative is positive for $x<1$ and negative for $x>1$ .

Answer

(1,e^{-1})

\text{maximum point}

Use the formula sheet

Every inside u contributes u′

eᵘ

→ u′eᵘ

always positive

ln u

→ u′ / u

u > 0

sin u

→ u′ cos u

radians

cos u

→ −u′ sin u

keep the minus

tan u

→ u′ sec²u

radians

Write u first when the inside is more than x.

The formula sheet supplies derivative rules. It does not identify the structure of the expression or add the chain-rule factor for you.

Keep nested rules separate

Worked example

A current-paper quotient

y=\frac{\ln(2x+1)}{x+3}

u'=\frac2{2x+1},\qquad v'=1

y'=\frac{\frac{2(x+3)}{2x+1}-\ln(2x+1)}{(x+3)^2}

The quotient rule earns the method mark; the derivative $2/(2x+1)$ is the chain-rule step inside it. (9709/22/M/J/24 Q6(a))

Form the stationary equation

Worked example

Exponential factor followed by a trig equation

y=e^x\cos x

y'=e^x(\cos x-\sin x)

Because $e^x\ne0$ , stationary points satisfy $\cos x-\sin x=0$ , so $\tan x=1$ . Now solve that trig equation over the stated interval.

Examiner note

Keep exact values unless the question asks for a decimal. If an angle must be approximated, solve over the full interval and round only the final answers.

Interpret the derivative sign

A derivative changing from positive to negative means the graph rises and then falls, so the turning point is a local maximum. A change from negative to positive gives a local minimum.

Your turnIndependent check [6 marks]

For $y=xe^{-x}$ , find the stationary point and use the sign of $dy/dx$ to determine its nature.

Show worked answer

y'=e^{-x}(1-x)

Since $e^{-x}>0$ , the derivative is positive for $x<1$ and negative for $x>1$ .

Answer

(1,e^{-1})

\text{maximum point}