Log derivatives

**Dr. Noobles** · November 29th 2006, 06:18 AM

Differenciate
1. F(x)= 5th root(lnx)
2. f(x)= sqroot(x) * ln(x)
3. F(t)= ln((2t+1)^3)/(3t-1)^4))

thanks for any help!

**Soroban** · November 29th 2006, 07:14 AM

Hello, Dr. Noobles!

$1)\;f(x)\:= \:\sqrt[5]{\ln x}$

We have: . $f(x)\;=\;\left(\ln x\right)^{\frac{1}{5}}$

Chain Rule: . $f'(x)\;=\;\frac{1}{5}\left(\ln x\right)^{-\frac{4}{5}}\cdot\frac{1}{x} \;= \;\frac{1}{5x}\left(\ln x\right)^{-\frac{4}{5}} \;=\;\frac{1}{5x\left(\ln x\right)^{\frac{4}{5}}}$

$2)\;f(x)\:= \:\sqrt{x}\cdot\ln x$

We have: . $f(x)\;=\;x^{\frac{1}{2}}\cdot\ln x$

Product Rule: . $f'(x)\;=\;x^{\frac{1}{2}}\!\cdot\!\frac{1}{x} + \frac{1}{2}x^{-\frac{1}{2}}\!\cdot\!\ln x \;= \;\frac{1}{x^{\frac{1}{2}}} + \frac{\ln x}{2x^{\frac{1}{2}}} \;= \;\frac{2 + \ln x}{2\sqrt{x}}$

$3)\;f(t) \:= \:\ln\left[\frac{(2t+1)^3}{(3t-1)^4}\right]$

Use log rules to simplify the function first . . .

$f(t) \:= \:\ln\left[\frac{(2t+1)^3}{(3t-1)^4}\right] \;=\;\ln(2t+1)^3 - \ln(3t-1)^4 \;=\;3\ln(2t+1) - 4\ln(3t-1)$

Then: . $f'(t)\;=\;3\!\cdot\!\frac{1}{2t+1}\!\cdot\!2 - 4\!\cdot\!\frac{1}{3t-1}\!\cdot\!3 \;=\;\frac{6}{2t+1} - \frac{12}{3t-1}$

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