ln Diff Problem

[Jason76]

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

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

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

\(\displaystyle f'(x) = [\ln 5x][6] + [\dfrac{30x}{x}] - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 30 - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 24\)

 

[Deleted member 4993]

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

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

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

\(\displaystyle f'(x) = [\ln 5x][6] + [\dfrac{30x}{x}] - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 30 - 6\)

\(\displaystyle f'(x) = 6 \ln 5x + 24\)

.

 

[HallsofIvy]

For problem like finding the derivative of ln(ax) there are several ways of thinking.

1) Let u= ax and use the chain rule: [d(ln(u))/du][du/dx]= [1/u][d(ax)/dx)]= [1/(ax)][a]= 1/x.

2) Same thing but done "on the fly": the derivative of ln(ax) is 1/(ax) multiplied by the derivative of ax, a: (1/ax)(a)= 1/x.

3) Easiest- using a property of logariths: ln(ax)= ln(a)+ ln(x). ln(a) is a constant so its derivative is 0. The derivative is 0+ 1/x= 1/x.