## - 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 $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}=-\left({3x\over x(x^2-3)^{{1\over2}}}\right) (x^2-3)$$
$$\displaystyle =-{3\over (x^2-3)^{{3\over2}}}$$