Product Rule ln

[Jason76]

\(\displaystyle f(x) = 6x \ln 5x - 6x\)

\(\displaystyle f'(x) = [\ln 5x][\dfrac{d}{dx} 6x] + [6x][\dfrac{d}{dx} \ln 5x] - \dfrac{d}{dx}6x\) - Product rule (on left) \(\displaystyle [g]['f] + [f]['g]\) given \(\displaystyle [f][g]\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{1}{u}(\dfrac{d}{dx} u)] - 6\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{1}{u}(5)] - 6\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{5}{u}] - 6\)\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{5}{u}] - 6\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{5}{5x}] - 6\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{1}{5x}] - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + \dfrac{6x}{x} - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 6 - 6\) What now?

 

[DrPhil]

\(\displaystyle f(x) = 6x \ln 5x - 6x\)

\(\displaystyle f'(x) = [\ln 5x][\dfrac{d}{dx} 6x] + [6x][\dfrac{d}{dx} \ln 5x] - \dfrac{d}{dx}6x\) - Product rule (on left)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{1}{5x}(5)] - 6\)

\(\displaystyle f'(x) = [\ln 5x][6] + [6x][\dfrac{1}{x}] - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + \dfrac{6x}{x} - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 6 - 6\) What now?

Do you remember enough algebra to subtract 6 from 6?

 

[Jason76]

Do you remember enough algebra to subtract 6 from 6?

I do, but the online homework says the answer is not \(\displaystyle 6\ln 5x\)

 

[DrPhil]

I do, but the online homework says the answer is not \(\displaystyle 6\ln 5x\)

I have checked your work again, and it still looks right. That suggests there is a typo somewhere, either in the question or in your entry of the answer. Do you need parentheses, 6 ln(5x) ?

 

[Jason76]

I have checked your work again, and it still looks right. That suggests there is a typo somewhere, either in the question or in your entry of the answer. Do you need parentheses, 6 ln(5x) ?

You were right. Now it says it's correct.