Calculus help?

[The_OG_Nelta]

i need help with calculus math... i am having trouble with chainrule combined with other rules... need major help

[Charla]

I see no problem for us to solve, so you must want links?
Try poking around here until someone shows up with a better one. Or just give us a problem to solve/explain while we complain about being used as a homework hotline.

[Jefe]

So wuts your q, niggra? Is it rules in general, or a specific problem you need help on? If it's the former, then just visit Paul's online Calculus, if it's the latter, just post the q.

Edit: Damn was late with the reply :O

[The_OG_Nelta]

Y = 2^sinpix

The answer won't help me, can anyone show me steps plz?

[Jefe]

Why are you using "y"? That implies implicit differentiation, is it?? I don't think so ;x

Anyways:
d/dx = (sin(pix))^2 = 2(Sin(pix)^1 (subtract one from the power 2, chain rule).
Next differentiate the base: 2(Sin(pix)^1(Cos(pix)),
then the derivative of the inner function:2(Sin(pix)^1(Cos(pix)(pi). So the answer is: 2pi(Sin(pix))(Cos(pix) = 0 I just woke up, so dunno if I got it right >_>

*Sin (pi) = 0, so when you multiply you get zeroth.

[Viena]

F(x) = 2^sin(pi*x)

f(x) = 2^x
g(x) = sin(x)
h(x) = pi*x

so:

F(x) = f(g(h(x)))

Let's derivate, apply the chain rule:

F'(x) = f'(g(h(x))) * (g(h(x)))'

Now, derivate g(h(x)), applying the chain rule again:

(g(h(x)))' = g'(h(x)) * h'(x)

Put it all together:

F'(x) = f'(g(h(x))) * g'(h(x)) * h'(x)

Now let's get f'(x), g'(x), h'(x):

f'(x) = (2^x) * ln(2)
g'(x) = cos(x)
h'(x) = pi

Do the substitution:

F'(x) = (2^(sin(pi*x))) * ln(2) * cos(pi*x) * pi

I hope this was clear enough, when you get used to it you can skip many of the steps.

[Charla]

It might be easier to think about if you call the parametric functions f(t), g(t) and h(t), to keep yourself from thinking they're all the real x.

[Norelco]

Just want to say I'm envious of all your math skillz. I aced all of my calculus classes in college but never used them since graduation so I barely remember any of it.

[Tricen]

I used to love math, I took college algebra and now I'm in statistics... Don't think I'll be taking anymore math.

[Demosthenes11]

can do it a much easier way than substitution

y = 2^sin(pi*x)
ln|y|=sin(pi*x)*ln(2) Take derivative here
1/y*dy=cos(pi*x)*pi*ln(2)*dx
dy/dx=cos(pi*x)*pi*ln(2)*y=cos(pi*x)*pi*ln(2)*2^sin(p i*x)

[Xajii]

can do it a much easier way than substitution

y = 2^sin(pi*x)
ln|y|=sin(pi*x)*ln(2) Take derivative here
1/y*dy=cos(pi*x)*pi*ln(2)*dx
dy/dx=cos(pi*x)*pi*ln(2)*y=cos(pi*x)*pi*ln(2)*2^sin(p i*x)

the guy needs help on the chain rule, i don't think using the dx rules for natural logs is going to make him any less confused at this point >_>

I would like to note that simply reading all of this instead of seeing it with the actual symbols makes it more difficult to understand, you should seek help IRL instead of online maybe >.>

[Jefe]

Oh shi, I thought it was (Sin(pix))^2 >__________________________________> Use parenthesis mofro :O

[Daahan]

I can't do problems like that without my Calculus notebook

[Viena]

the guy needs help on the chain rule, i don't think using the dx rules for natural logs is going to make him any less confused at this point >_>

I would like to note that simply reading all of this instead of seeing it with the actual symbols makes it more difficult to understand, you should seek help IRL instead of online maybe >.>

Ditto. Let's ask for a LaTeX mod!

[Olo401]

OMG.. yallaremeltinmybrainwithurevylcalculusthreadkthxdi e!

[kuronosan]

Calculus.

It says you + me = us.