- Differentiation of the Natural Logarithm

\(\displaystyle y=\ln ⁡x \to y'={1\over x}\)

\(\displaystyle y=\ln⁡(u(x)) \to y'={u' (x) \over u(x)}\)

LESSON 1
Differentiate each of the following.
(a) \(y=\ln⁡(3x+1)\)
(b) \(y=3 \ln⁡(7x-2)\)
(c) \(y=\ln⁡(x^2-x+1)\)
(d) \(y=\ln⁡(\sin ⁡4x)\)
(e) \(y=x^2 \ln ⁡x\)
(f) \(\displaystyle y={2 \ln ⁡x\over 1+x}\)
(g) \(y=\ln⁡[(2x-1)(3x+2)^2]\)
SOLUTION
(a) \(y=\ln⁡(3x+1)\)
\(u(x)=3x+1 → u'(x)=3\)
\(\displaystyle {dy\over dx}={3\over 3x+1}\)

(b) \(y=3 \ln⁡(7x-2)\)
\(u(x)=7x-2 → u'(x)=7\)
\(\displaystyle {dy\over dx}=3\left({7\over 7x-2}\right)\)
\(\displaystyle {dy\over dx}={21\over 7x-2}\)

(c) \(y=\ln⁡(x^2-x+1)\)
\(u(x)=x^2-x+1 → u'(x)=2x-1\)
\(\displaystyle {dy\over dx}={2x-1\over x^2-x+1}\)

(d) \(y=\ln⁡(\sin⁡ 4x)\)
\(u(x)=\sin⁡4x → u'(x)=4 \cos⁡4x\)
\(\displaystyle {dy\over dx}={4 \cos⁡4x\over \sin⁡4x}\)
\(=4 \cot⁡4x\)

(e) \(y=x^2 \ln ⁡x\)
    • Use the PRODUCT RULE
\(\displaystyle {dy\over dx}=2x \ln ⁡x+x^2 \left({1\over x}\right)\)
\(=2x \ln⁡ x+x\)
\(=x(2 \ln ⁡x+1)\)

(f) \(\displaystyle y={2 \ln ⁡x\over 1+x}\)
    • Use the QUOTIENT RULE
\(\displaystyle {dy\over dx}=\left(2\times {1\over x}\right)(1+x)-{2 \ln ⁡x (1)\over (1+x)^2}\)
\(\displaystyle ={2(1+x)-2x \ln⁡ x \over x(1+x)^2}\)
\(\displaystyle ={2+2x-2x \ln ⁡x)\over x(1+x)^2}\)
\(\displaystyle ={2(1+x-x \ln ⁡x)\over x(1+x)^2}\)

(g) \(y=\ln⁡[(2x+1)(3x+2)^2]\)
    • Use the properties of logarithms, \(\log ⁡XY=\log ⁡X+\log ⁡Y\) and \(\log⁡ X^n=n \log ⁡X\)
\(y=\ln⁡(2x+1)+\ln⁡(3x+2)^2\)
\(y=\ln⁡(2x+1)+2 \ln⁡(3x+2)\)
\(\displaystyle {dy\over dx}={2\over 2x+1}+2\left({3\over 3x+2}\right)\)
\(\displaystyle ={2\over 2x+1}+{6\over 3x+2}\)
\(\displaystyle ={2(3x+2)+6(2x+1)\over (2x+1)(3x+2)}\)
\(\displaystyle ={6x+4+12x+6\over (2x+1)(3x+2)}\)
\(\displaystyle ={2(9x+5)\over (2x+1)(3x+2)}\)

LESSON 2
Differentiate \(3^x\) w.r.t \(x\)
SOLUTION
  • Let \(y=3^x\)
  • Take ln of both sides
\(\ln⁡ y=\ln⁡ 3^x\)
  • Use implicit differentiation
\(\displaystyle {1\over y} {dy\over dx}=x \ln⁡3\)
  • Solve for \(\displaystyle {dy\over dx}\)
\(\displaystyle {dy\over dx}=y \ln⁡3\)
\(=3^x \ln⁡3\)
In general, if \(y=a^x\) then \(\displaystyle {dy\over dx}=a^x \ln⁡ a\)

LESSON 3
If \(\displaystyle y={x\over \sqrt{x^2-3}}\), find \(\displaystyle {dy\over dx}\)
SOLUTION
\(\displaystyle y={x\over \sqrt{x^2-3}}\)
  • Take ln ⁡of both sides
\(\displaystyle \ln ⁡y=\ln⁡ \left({x\over \sqrt{x^2-3}}\right)\)
  • Apply the logarithmic property: \(\displaystyle \log⁡\left({X\over Y}\right)=\log ⁡X-\log ⁡Y\)
\(\displaystyle \ln ⁡y=\ln ⁡x-\ln⁡(x^2-3)^{{1\over2}}\)
  • Use implicit differentiation
\(\displaystyle {1\over y} {dy\over dx}={1\over x}-{1\over2} \left({2x\over x^2-3}\right)\)
\(\displaystyle ={1\over x}-{x\over x^2-3}\)
\(\displaystyle ={x^2-3-x^2\over x(x^2-3)}\)
\(\displaystyle =-{3\over x(x^2-3)}\)
  • Solve for \(\displaystyle {dy\over dx}\)
\(\displaystyle {dy\over dx}=-{3y\over x(x^2-3)}\)
  • Sub \(\displaystyle y={x\over (x^2-3)^{{1\over2}}}\) into \(\displaystyle {dy\over dx}\)
\(\displaystyle {dy\over dx}=-\left({3x\over x(x^2-3)^{{1\over2}}}\right) (x^2-3) \)
\(\displaystyle =-{3\over (x^2-3)^{{3\over2}}}\)