Question

63 - 74. derivatives of logarithmic functions calculate the derivative of the following functions. in some cases, it is useful to use the properties of logarithms to simplify the functions before computing f(x).

1. y = 4 log₃(x² - 1)
2. y = log₁₀x
3. y = (cos x)ln cos²x
4. y = log₈|tan x|
5. y = 1/log₄x
6. y = log₂(log₂x)
7. f(x) = ln(3x + 1)⁴
8. f(x) = ln(2x/(x² + 1)³)

Explanation:

1. For \(y = 4\log_3(x^2 - 1)\):

• Recall the change - of - base formula \(\log_a u=\frac{\ln u}{\ln a}\) and the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\).
• First, rewrite \(y = 4\frac{\ln(x^2 - 1)}{\ln 3}\).
• Then, find the derivative. The derivative of \(\ln u\) with respect to \(x\) is \(\frac{u^\prime}{u}\). Here \(u = x^2-1\), so \(u^\prime = 2x\).
• \(\frac{dy}{dx}=\frac{4}{\ln 3}\cdot\frac{2x}{x^2 - 1}=\frac{8x}{(x^2 - 1)\ln 3}\).

1. For \(y=\log_{10}x\):

• Using the change - of - base formula \(\log_{10}x=\frac{\ln x}{\ln 10}\).
• The derivative of \(\ln x\) is \(\frac{1}{x}\), so \(\frac{dy}{dx}=\frac{1}{x\ln 10}\).

1. For \(y = (\cos x)\ln(\cos^2x)\):

• First, use the property \(\ln a^b = b\ln a\) to rewrite \(y = 2(\cos x)\ln(\cos x)\).
• Then, use the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = 2\cos x\) and \(v=\ln(\cos x)\).
• \(u^\prime=-2\sin x\), and for \(v=\ln(\cos x)\), using the chain rule, \(v^\prime=\frac{-\sin x}{\cos x}=-\tan x\).
• \(\frac{dy}{dx}=-2\sin x\ln(\cos x)+2\cos x\cdot(-\tan x)=-2\sin x\ln(\cos x)-2\sin x=-2\sin x(1 + \ln(\cos x))\).

1. For \(y=\log_8|\tan x|\):

• Using the change - of - base formula \(y=\frac{\ln|\tan x|}{\ln 8}\).
• The derivative of \(\ln|\tan x|\) using the chain rule: The derivative of \(\tan x\) is \(\sec^2x\), so the derivative of \(\ln|\tan x|\) is \(\frac{\sec^2x}{\tan x}=\frac{1}{\sin x\cos x}\).
• \(\frac{dy}{dx}=\frac{1}{\ln 8}\cdot\frac{\sec^2x}{\tan x}=\frac{1}{\sin x\cos x\ln 8}\).

1. For \(y=\frac{1}{\log_4x}\):

• Using the change - of - base formula \(\log_4x=\frac{\ln x}{\ln 4}\), so \(y = \frac{\ln 4}{\ln x}\).
• Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\), with \(u = \ln 4\) (a constant, so \(u^\prime = 0\)) and \(v=\ln x\) (\(v^\prime=\frac{1}{x}\)).
• \(\frac{dy}{dx}=-\frac{\ln 4}{x(\ln x)^2}\).

1. For \(y=\log_2(\log_2x)\):

• Using the change - of - base formula \(\log_2u=\frac{\ln u}{\ln 2}\), so \(y=\frac{\ln(\ln x)}{\ln 2}\).
• Using the chain rule, the derivative of \(\ln(\ln x)\) is \(\frac{1}{x\ln x}\).
• \(\frac{dy}{dx}=\frac{1}{x\ln x\ln 2}\).

1. For \(f(x)=\ln(3x + 1)^4\):

• Using the property \(\ln a^b = b\ln a\), \(f(x)=4\ln(3x + 1)\).
• Using the chain rule, the derivative of \(\ln(3x + 1)\) is \(\frac{3}{3x+1}\), so \(f^\prime(x)=\frac{12}{3x + 1}\).

1. For \(f(x)=\ln\frac{2x}{(x^2 + 1)^3}\):

• Using the property \(\ln\frac{a}{b}=\ln a-\ln b\), \(f(x)=\ln(2x)-3\ln(x^2 + 1)\).
• The derivative of \(\ln(2x)=\frac{1}{x}\) and the derivative of \(\ln(x^2 + 1)\) using the chain rule is \(\frac{2x}{x^2 + 1}\).
• \(f^\prime(x)=\frac{1}{x}-\frac{6x}{x^2 + 1}=\frac{x^2 + 1-6x^2}{x(x^2 + 1)}=\frac{1 - 5x^2}{x(x^2 + 1)}\).

Step1: Recall derivative rules

Recall change - of - base formula \(\log_a u=\frac{\ln u}{\ln a}\), chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), product rule \((uv)^\prime=u^\prime v+uv^\prime\), quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\) and \(\ln a^b = b\ln a\), \(\ln\frac{a}{b}=\ln a-\ln b\).

Step2: Rewrite functions

Rewrite each logarithmic function using properties of logarithms to simplify before differentiation.

Step3: Differentiate

Apply appropriate derivative rules to find the derivative of each function.

Answer:

• For \(y = 4\log_3(x^2 - 1)\), \(y^\prime=\frac{8x}{(x^2 - 1)\ln 3}\)
• For \(y=\log_{10}x\), \(y^\prime=\frac{1}{x\ln 10}\)
• For \(y = (\cos x)\ln(\cos^2x)\), \(y^\prime=-2\sin x(1+\ln(\cos x))\)
• For \(y=\log_8|\tan x|\), \(y^\prime=\frac{1}{\sin x\cos x\ln 8}\)
• For \(y=\frac{1}{\log_4x}\), \(y^\prime=-\frac{\ln 4}{x(\ln x)^2}\)
• For \(y=\log_2(\log_2x)\), \(y^\prime=\frac{1}{x\ln x\ln 2}\)
• For \(f(x)=\ln(3x + 1)^4\), \(f^\prime(x)=\frac{12}{3x + 1}\)
• For \(f(x)=\ln\frac{2x}{(x^2 + 1)^3}\), \(f^\prime(x)=\frac{1 - 5x^2}{x(x^2 + 1)}\)