can you take derivative of this using chain rule?

Is there anything more disappointing than coming here and seeing people posting high school math shit?

I'm not sure about this but I guess it has something to do with the chain rule. (https://en.wikipedia.org/wiki/Chain_rule)

When you want to differentiate your first function you're actually substituting the function in the exponential by a new function (let's call it u(x)) so now your derivative becomes:
f(x)=exp(x^2+x), u(x)=x^2+x
df(x)/dx=df/du*du/dx

In your second function there is no function of x to substitute in your exponential. Therefore the derivative becomes 4*x^3.

>>7686733
Wrong. There is the function u(x) = 4.

The point is that you'd be differentiating something of the form x^{u(x)} which is quite different from an exponential, or a polynomial. So you'd have
[math]y = x^{u(x)}[/math] gives [math]y'/y = (u(x)\log(x))'[/math].

I'll let you work out the details to check it all works.

>>7686737
Ah I see. So what OP is doing is confusing different techniques of differentiation?

>>7686742
Not at all, just differentiation incorrectly. If one wants to treat the exponent of 4 as a function of x (perfectly fine), one has to remember that the base is also a function of x, and therefore lends itself to be better computed via implicit differentiation. Using the power rule or exponential rule is simply incorrect.

>>7686737
I am op and i didnt understand anything by your answer, i dont know logarithm, so how is it done using chain rule?

>>7686746
please bro i am about to go crazy please is there no other way for you to put simplier?

>>7686810
I know implicit diff. but just couldnt see what you are doing...

different anon here

>>7686749
What that anon means is that you are mis-using chain rule there and to use the chain rule correctly is quite a bit more complicated than what you are suggesting.

First, remember that all of these differentiation rules come from the limit process. By using that, you can clearly see that d/dx(x^4) = 4x^3. And, from the same limit process, we can generalise for the case of x^c where c is a constant term. (And only when c is a constant!)

But x^(something that isn't a constant) is a different thing altogether. For example:

d/dx(x^x) is not equal to x*x^(x-1)
which would be mis-using the general rule for x^c (which we can't do because x is not a constant)

And, if you want to differentiate x^4 using chain rule (which is not a particularly efficient way to do it), then you need to use something that gets called the Functional Power Rule.

So a function to the power of a function (f^g) is not the same thing as exp^f or f^2, and therefore it needs to be treated differently. And you have to know about logs... and implicit differentiation. The formula you want is, (written in a different form to the other anon):

If y = u^v
Then
dy/dx = (u^v)*(lnu*(dv/dx) + (v/u)*(du/dx))

The good news here is that because d/dx(4)=0, you don't have to worry about whatever "lnx" means to see that your differentiation works.

>>7686847
Thanks bro, you are a true guy. We don't do it because it is something different. I am going to take a look at this after some time when i know about logs.

how about diffrentiating sinx using chain rule? Am i correct like this?

1.sinx^0 (comes from (sinx)^1)

cosx (comes from sinx)

1 (comes from dx/dx)

1*sinx^0*cosx*1=cosx?

>>7686862
Not exactly sure what you've got there...

Is your chain rule substitution u=sinx?
Then
d/dx(sinx) = (du/dx)*d/du u
= (d/dx sinx) * 1
= cos x

I don't think a sinx^0 really comes in to it...

Or maybe like this:

u = x
so
d/dx sinx = (du*dx)*(d/du sin u)
= (d/dx x)*cos u
= 1* cos x
= cos x

Both of these exercises are fairly redundant because I've had to use knowledge of the derivative of sinx to find the answer (which is the derivative of sinx).

Good luck with it anyway.

File: Lain_In_Fetal.gif (20KB, 387x527px) Image search: [Google]

20KB, 387x527px

>do the short hand of a proof
>WOW WHY DOES IT NOT WORK FOR THIS SITUATION

>>7686883
Oops

That second one should be
u = x
so
d/dx sinx = (du/dx)*(d/du sin u)
= (d/dx x)*cos u
= 1* cos x
= cos x

>>7686885
Welcome to /sci/ anon :^(

>>7686883
1*sinx^0 comes because just as why we do x^2 as 2x.

>when we take derivative of e^(x^2+x), we treat x^2+x as a function.
yes the inside function
>But how can you take derivative of x^4 using chain rule?
you're supposed to use the power rule

File: 1448744066294.gif (20KB, 387x527px) Image search: [Google]

20KB, 387x527px

>>7686885
I could not help but notice your gif was not optimized anon.
I have optimized your gif.
Your gif is now optimized.

>>7686813
Original Anon here, and yes you'll need to understand logarithms for the details I had above, but the general concept you can grasp already.

The chain rule is about composition of functions. It works (it better!) even when you have a constant function, like u(x) = 4. But in order to use the chain rule, you have to understand how to differentiate things like [math]x^{u(x)}[/math]. You can just accept at this point you don't have a method to handle this type of function (the answer comes as above, with logarithms and implicit differentiation, something to learn in the future).

For example, how would you differentiate [math]x^x[/math]? Not with the power rule, and not like an exponential. It's just something different you'll learn to handle. If you want to use the chain rule on [math]x^4[/math], you're in this mixed case. I hope you can at least see that.

>>7686737
But you're not differentiating with respect to 4, you're differentiating with respect to x.

If you were differentiating with respect to 4, you would be right to compute d/dx of the exponent in applying the chain rule.

>>7686722
Because there's no x to derivate in that power, you can't actually derivate unless you do d(4) instead of dx, which will always be 0 anyway.

>>7688636
When did I differentiate "with respect to 4"? I wrote it as a function of x, the variable I'm differentiating with respect to, then performed the usual differentiation techniques.

>>7688744
But you are differentiating woth respect to u, sp the answer is going to be 0, obviously, since u is a constant.

>>7688744
>>7688760
Btw, you can't x^u(x) with u = 4. It doesn't make sense.

errr guys....OP's question confused me.
if f(x)=x^u(x), then by the chain rule
df/dx=(df/du)*(du/dx)
but if u(x)=constant then du/dx=0 and df/dx=0 but we know this isn't true....
Where am I going wrong?

>>7688769
I think I may have divided by zero here.

>>7688769
Because x^u(x) is wrong, it should be x^u(y).

>>7688777
No I don't think that's the case, u can be a function of whatever we want it to be.

>>7688785
Test it.

>>7688791
Alright
df/dx=(df/du)(du/dy)(dy/dx)
which is still zero.

>>7688795
It should be 0, did you expect anything else?

>>7688799
Yes, if u(y)=2 then f(x)=x^2 so df/dx = 2x
I think the problem here is this:
df/dx=(df/du)*(du/dx)=(du/dx)/(du/df) and for u=constant this is df/dx=0/0

>>7688804
But in principle it doesn't make any sense because you are differentiating woth respect to x so it's df/dx = ux. There are no more steps to it unless you are differentiating with respect to u to begin with.

>>7688767
Because it's u(x) = 4... I'm not sure what you're misunderstanding. Do you believe f(x) = c is a function of x where c is a constant?

>>7686722
If g(x)=4 then f(g(x))=f(4)!=x^4.

>>7688926
You can expand this to g being any constant obviously.

>>7688744
The chain rule assumes a composition of functions. The fact that you even added the "*0" term already implies you're deriving with respect to whatever term is in the exponent.

>>7689622
Are we disagreeing? I'm not quite sure what the confusion is anymore.