Ln derivative. 242.7.3.83

[karush]

$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right]$$
Assume first step is expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right)$$

 

[Greg]

Re: Ln integral. 242.7.3.83

Yes. Now use

$$\dfrac{d\ln(f(x))}{dx}=\dfrac{f'(x)}{f(x)}$$

 

[HallsofIvy]

Re: Ln integral. 242.7.3.83

Why was this titled "Ln integral" when the problem asks for a derivative?

 

[karush]

$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right] \ \ \ \ \
f'(x)=\frac{3\left(4{x}^{2}-16x-45 \right)}{(x+6)(2x-1)(2x+3)}$$
expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right) $$
then
$$\dfrac{f'(x)}{f(x)}
= \frac{2}{2x+3}+\frac{5}{x+6}-\frac{6}{2x-1}$$$$$$

 

[topsquark]

$\large{242.7.3.83}$

Differentiate
$$\displaystyle
f(x)=\ln\left[{\frac{(2x+3)(x+6)^5}{(1-2x)^3}}\right] \ \ \ \ \
f'(x)=\frac{3\left(4{x}^{2}-16x-45 \right)}{(x+6)(2x-1)(2x+3)}$$
expansion..
$$f(x)=\ln\left({2x+3}\right)
+5\ln\left({x+6}\right)
-3\ln\left({1-2x}\right) $$
then
$$\dfrac{d\ln(2x+3)}{dx}=\dfrac{f'(x)}{f(x)}$$

In this case, since you are doing this one term at a time, f(x) = 2x + 3. Then for the second term f(x) = x + 6 and for the third f(x) = 1 - 2x

-Dan