# using diff func of a func..get from this to this???!!!!

Discussion in 'Undergraduate Math' started by Electromechy1, Sep 20, 2003.

1. ### Electromechy1Guest

hi guys,
came across this one in book i'm using, its in the diff func of a func section
so no cheating using the standard derivative e^ax=ae^ax. how do i wtite this in
standard form so i can use func of a func?? knowing my luck will come up on
exam, so a solution would be appreciated.

5e^2t+1 to 10e^2t+1

best regards

shaun

Electromechy1, Sep 20, 2003

2. ### David MoranGuest

I'll rewrite it as y=(5e^2t+1)^(10e^2t+1)

First take the log of both sides and use the Power Rule.
ln y=(10e^2t+1)ln(5e^2t+1)

Now differentiate implicitly
(1/y)(dy/dx)=(10e^2t+1)(10e^2t/(5e^2t+1))+(10e^2t)ln(5e^2t+1)
dy/dx=(5e^2t+1)^(10e^2t+1)((10e^2t+1)(10e^2t/(5e^2t+1))+(10e^2t)ln(5e^2t+1)

David Moran

David Moran, Sep 20, 2003

3. ### Paul SperryGuest

?

What is "this"?

Paul Sperry, Sep 21, 2003
4. ### Electromechy1Guest

thanx david

regards

shaun

Electromechy1, Sep 21, 2003
5. ### Jonathan MillerGuest

Alternatively, you could just use the chain rule. You've got y = f(x)^g(x)
=e^[ln(f(x))*g(x)], so
dy/dx = e^[ln(f(x))*g(x)]*[1/f(x)*f'(x)*g(x)+ln(f(x))*g'(x)], and plug in
what f(x) and g(x) are.

Whatever you're more comfortable with. Objectively, it looks like six of
one and half a dozen of the other to me.

I was going to suggest that you need to do it both ways and compare the
answers, but it's likely that it'll just be real hard to chase through all
the algebra, and you'll be back for us to check your work. That's too ugly,
so just take my word that the answers are really the same.

Jon Miller

Jonathan Miller, Sep 21, 2003