Composite Logarithmic Functions | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Inside derivative over inside

Key idea

\frac{d}{dx}(\ln u)=\frac{u'}u

, for

u>0

Worked example

Differentiate a logarithm of a linear function

y=\ln(3x-2)

y'=\frac3{3x-2},\qquad x>\frac23

Common mistake

The derivative is not

1/(3x-2)

. The numerator 3 comes from differentiating the inside.

Use logarithm laws

If every logarithm is defined, then $\ln(ab)=\ln a+\ln b$ , $\ln(a/b)=\ln a-\ln b$ and $\ln(a^k)=k\ln a$ . These identities can replace a longer product or quotient inside the logarithm.

Worked example

Simplify the logarithm first

Work on the interval $x>-1/2$ . This makes both $2x+1$ and $x+4$ positive, so each logarithm below is defined.

y=\ln\left(\frac{(2x+1)^3}{x+4}\right)

\begin{aligned}y={}&3\ln(2x+1)\\&-\ln(x+4)\end{aligned}

y'=\frac6{2x+1}-\frac1{x+4}

Examiner note

\ln((2x+1)/(x+4))

and

\ln(2x+1)/(x+4)

are different functions. Parentheses decide whether the quotient is inside or outside the logarithm.

Combine rules

Your turnIndependent check [5 marks]

Differentiate $y=x^2\ln(3x)$ . State the domain.

Show worked answer

\frac{d}{dx}\ln(3x)=\frac3{3x}=\frac1x

y'=2x\ln(3x)+x^2\left(\frac1x\right)

Answer

y'=2x\ln(3x)+x

x>0

Inside derivative over inside

Key idea

\frac{d}{dx}(\ln u)=\frac{u'}u

, for

u>0

Worked example

Differentiate a logarithm of a linear function

y=\ln(3x-2)

y'=\frac3{3x-2},\qquad x>\frac23

Common mistake

The derivative is not

1/(3x-2)

. The numerator 3 comes from differentiating the inside.

Use logarithm laws

If every logarithm is defined, then $\ln(ab)=\ln a+\ln b$ , $\ln(a/b)=\ln a-\ln b$ and $\ln(a^k)=k\ln a$ . These identities can replace a longer product or quotient inside the logarithm.

Worked example

Simplify the logarithm first

Work on the interval $x>-1/2$ . This makes both $2x+1$ and $x+4$ positive, so each logarithm below is defined.

y=\ln\left(\frac{(2x+1)^3}{x+4}\right)

\begin{aligned}y={}&3\ln(2x+1)\\&-\ln(x+4)\end{aligned}

y'=\frac6{2x+1}-\frac1{x+4}

Examiner note

\ln((2x+1)/(x+4))

and

\ln(2x+1)/(x+4)

are different functions. Parentheses decide whether the quotient is inside or outside the logarithm.

Combine rules

Your turnIndependent check [5 marks]

Differentiate $y=x^2\ln(3x)$ . State the domain.

Show worked answer

\frac{d}{dx}\ln(3x)=\frac3{3x}=\frac1x

y'=2x\ln(3x)+x^2\left(\frac1x\right)

Answer

y'=2x\ln(3x)+x

x>0