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[frostshotgg]

The derivative of e^(x) is e^(x). Does this apply to any constant^(x) or just the constant e?

Like, would derivative of 2^(x) be 2^(x)?

e is a very special number. The fact that f'(x) = f(x) in that special case is one of the many definitions of e. There is no other function with a derivative that equals itself, although there is another group of functions that can eventually wind up with what you started, the trig functions.

[Spehss _]

New question, if I need to find y' of y=2^(sin(πx) using the Chain rule. (note that π = pi) Can I use the chain rule just once or do I need to use it multiple times?

As far as I can see, either y=f(g(x)) where f(x)=2^(sin(x)) and g(x)=πx or y=f(g(h(x))) where f(x)=2^x and g(x)=sin(X) and h(x)=πx.

[frostshotgg]

He hasn't gotten to implicit differentiation yet, I would assume given that he's only just not learning dy/dx(e^x) and chain rule. His teacher is almost certainly looking for the standard differentiation method.

Chain rule has to be applied for every single function. In that case, your equation can be broken down into 3 distinct functions that you must differentiate sequentially.
y = f(g(h(x))), f(x) = 2^x, g(x) = sin(x), h(x) = pi*x. To apply chainrule, you work sequentially outside in.

For f(x) = 2^x, you have a form of a^x, whose derivative rule is d(a^x )/dx = a^x * ln(a). In this case, that becomes 2^"x" * ln(2). I single out the x there, because it's important to remember that your "x" there is actually g(h(x)), which is sin(pi*x), so your current form becomes 2^(sin(pi*x)) *ln(2). Then you do chain rule.

For g(x) = sin(x), d/dx = cos(x), no transformation needed. So now you multiply that with your f(x). 2^(sin(pi*x)) * ln(2) * cos("x"). Again, remember that "x" in what you just added is h(x), so pi*x. Your total form is now 2^(sin(pi*x)) * ln(2) * cos(pi*x). Then you do chain rule again.

For h(x) = pi*x, d/dx = pi. This is a form of d(ax)/dx = a. You should be able to recognize that a constant times x's derivative is always the constant. So then you multiply in that derivative, and you get your final answer, 2^(sin(pi*x)) * ln(2) * cos(pi*x) * pi.

You can check this answer by putting it into your TI-89's derivative function, or on an online calculator such as WolframAlpha, like this. If you did everything right, what you have should be equivalent to that. In this case, your ln(2) and pi are shuffled around which is fine thanks to commutative property of multiplication.

EDIT: Interesting. I learned a new bbcodes shorthand today.

[TheBiggerFish]

Can you do derivatives (or whatever frostshotgg said to do with a TI-89) on a TI-84?

[frostshotgg]

Unfortunately not. If you're taking calculus now, and expect to take another calculus course any time ever, a TI-89 or a TI-nSpire is an invaluable resource and worth the money. They have derivative and integral functions that are incredibly useful, as well as other useful functions, like summations, factorization, and especially equation solvers. The other ones are useful, but the equation solving function is just so fantastic that I can't imagine talking calculus without it.

[TheBiggerFish]

...I wonder if "Calculator says so" is acceptable showing of work, in any case.
Or if you can implement something in Pascal...
But no has monies so
Calc hurts.
We're doing limit definition of derivatives RN.

[frostshotgg]

I would ask your school if they have TI-89s to loan out, then. If they don't, then it's probable they won't have any questions that require a calculator.

Generally speaking, if you're allowed a calculator on the test/problem, just writing d/dy (f(x)) = whatever is sufficient.

[TheBiggerFish]

Hah, no.  We have to buy calculators.
...By the way, are those off-limits on like the ACT and stuff?

[frostshotgg]

TI-89 and other calculators with equation solving functions are off limits on the ACT specifically because they can solve equations. They're fair game pretty much everywhere else, and the AP Calculus tests basically require a calculator as smart as the 89 for the calculator allowed portions.

[TheBiggerFish]

Yeah, I figured as much.  Oh well, I'll put it on my List Of Things That I Would Like For College.

[Spehss _]

I swear I'm doing all the problems wrong holy shit.

Get the derivative of y=2/(1+e^(-x))

So y=f(g(h(x)))

f(x)=2/x where x=g(h(x))

g(x)=1+x where x=h(x)

h(x)=e^(-x)

Am I doing it right so far?

Textbook describes chain rule for 3 links as (dy/dt)=(dy/dx)(dx/dt)=(dy/dx)(du/dx)(dx/dt). No idea how to interpret that. I know think I know as far as I know that d(insert function here)/dx refers to "derivative of function in the numerator" or some shit. Don't know what the "dx" or "dt" in the denominator of the derivative symbolizes or how I should interpret that as anything aside from "derivative of function in the numerator".

So (dy/dx)(du/dx)(dx/dt) = "derivative of y" * "derivative of u" * "derivative of x" ? Does this mean that to find y' I need to multiply f'(x) by g'(x) by h'(x)?

My notes for a chain rule with 2 links is y'=f'(g(x))*g'(x). Seems like 3 links would be y'=f'(g(h(x)))*g'(x)*h'(x). Or y'=f'(g(h(x)))*g'(h'(x))

...I don't know what I'm doing.

[Graknorke]

Derive y in terms of x (but you probably meant u otherwise it doesn't make much sense), u in terms of x, and x in terms of t.

[Graknorke]

Oh right, looking at the question that should come out quite easily. I retract any endorsement of using the chain rule.

[Spehss _]

I think you're overthinking this one. Consider just using the quotient rule. Save yourself a lot of headache.

Quotient rule is a thing the class hasn't covered yet, thanks. Otherwise I would clearly use the easiest solution.

[Helgoland]

It's a special case of the chain rule anyway.