Basic Calculus Homework: How to Differentiate 6e^(-t).ln(t+5) - Check My Work!

[Patjamet]

It's a basic calculus problem for some uni homework, but it's been a while and I'm rusty.
More or less just looking for a check on my work

Homework Statement

Differentiate:

v(t)= 6e^(-t).ln(t+5)

The Attempt at a Solution

Using the product rule:

v'(t) = du/dt(6e^(-t)).ln(t+5) + 6e^(-t).dv/dt(ln(t+5))

v'(t) = -6e^(-t).ln(t+5) + 6e^(-t).1/(t+5)

v'(t) = -6e^(-t).ln(t+5) + 6e^(-t)/(t+5)

Is this correct?

Thanks,

Pat.

 

[Encephalon]

Looks right to me!

 

[Rasalhague]

Yes, it's correct. But your notation is a little odd in the first line. Instead of

v'(t) = du/dt(6e^(-t)).ln(t+5) + 6e^(-t).dv/dt(ln(t+5))

you could let f(t)=6e^(-t), and g(t)=ln(t+5), then simply write

v'(t)=(fg)'(t) = f'(t) \cdot g(t) + f(t) \cdot g'(t)

or you could use the d/dt notation like this

\frac{\mathrm{d}v}{\mathrm{d} t} = \frac{\mathrm{d}fg}{\mathrm{d} t} = \frac{\mathrm{d}f}{\mathrm{d} t} \cdot g + f \cdot \frac{\mathrm{d} g}{\mathrm{d} t}

or

\frac{\mathrm{d}}{\mathrm{d} t} (6e^{-t}) \bigg|_{t=t_0} \cdot \ln (t_0+5)+6e^{-t_0} \cdot \frac{\mathrm{d} }{\mathrm{d} t}(\ln(t+5)) \bigg|_{t=t_0}

or

\frac{\mathrm{d} (6e^{-t})}{\mathrm{d} t} (t_0) \cdot \ln (t_0+5)+6e^{-t_0} \cdot \frac{\mathrm{d} (\ln(t+5))}{\mathrm{d} t} (t_0)

or with some such variation.

By the way, if you just want to check a result, there are many online calculators; here's a good one:

http://www.wolframalpha.com

 

[Patjamet]

Thanks for your replies guys.

Rasalhague, thanks for your corrections I'll make note of them in my studies

 

[Rasalhague]

You're welcome, Patjamet. I should add that you'll also see variations of the d/dt (Leibniz) notation where the same letter, t, is used both for the variable in the name of the function, and for a particular value of it. If you're familiar with the concepts, it's usually easy to disambiguate, but when learning new ideas, I find it helps to keep things as clear as possible.

Also, the Leibniz notation is often used ambiguously, so, people may write y(t) = y = 6e^-t, and y' = y'(t) = dy/dt, where it's not clear whether dy/dt denotes the function or the value of this function. Again, when performing familiar calculations, it may not matter, but for understanding new concepts, it's good to have a notation that keeps these things distinct, or at least to be have an unambiguous notation to fall back on if things get confusing.

Another notation I've seen, which is nicely simple, is Df for the derivative as a function, f', and Df(t) for its value, f'(t), at a particular point, t.

 

[crewlah]

Strange - I have the exact question for a calculus assignment and I got stuck on the same part as Patjamet - Thanks for the clarification Rasal!