differentiation ... help

[bimbo36]

 

 

 

where do i start ?

Edited March 31, 2011 by bimbo36

[Hal.]

Help not answers ! if y = ( sin x ) ^ 0.5 then the derivative is 0.5 ( ( sin x ) ^ - 0.5 ) ( cos x ) = 0.5 ( cos x ) / ( ( sin x ) ^ 0.5 )

 

 

 

 

[mooeypoo]

This is a chain rule problem. Go from "outwards" and then "inwards" in sequence.

 

[math]y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx}}}[/math]

 

Start with the differentiation of the big square root, then multiply by the derivative inside it, which includes a square root, etc.

 

If you post how you started it, we can help you continue. We don't give final answers here, just help you get there.

 

~mooey

[ydoaPs]

This is a chain rule problem. Go from "outwards" and then "inwards" in sequence.

 

[math]y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx}}}[/math]

 

Start with the differentiation of the big square root, then multiply by the derivative inside it, which includes a square root, etc.

 

If you post how you started it, we can help you continue. We don't give final answers here, just help you get there.

 

~mooey

That's what I thought at first, but it might be infinitely nested.

[math]y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+...}}}[/math]

 

Or maybe not. It's written in an odd way.

Edited March 31, 2011 by ydoaPs

[mooeypoo]

Are the ellipses (...) inside or outside of the root? They look to be outside of it...

 

The original poster should answer that, though....

[ydoaPs]

Are the ellipses (...) inside or outside of the root? They look to be outside of it...

 

The original poster should answer that, though....

That's what I'm saying. If they're outside, it's not terribly difficult at all.

 

They should have written it differently. Perhaps:

In the following problems, find dy/dx:

1)

 

[math]y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx}}}[/math]

[mooeypoo]

That's what I'm saying. If they're outside, it's not terribly difficult at all.

 

They should have written it differently. Perhaps:

 

Just out of personal experience, this looks like a screenshot from a list of questions with instructions. I am guessing it's the simpler version.

I await bimbo36 to clarify.

 

~mooey

 

 

[bimbo36]

This is a chain rule problem. Go from "outwards" and then "inwards" in sequence.

 

[math]y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx}}}[/math]

 

Start with the differentiation of the big square root, then multiply by the derivative inside it, which includes a square root, etc.

 

If you post how you started it, we can help you continue. We don't give final answers here, just help you get there.

 

~mooey

 

lets see ..

 

a single y is sinx+sinx+sinx

 

The definition of the derivative can be approached in two different ways. One is geometrical (as a slope of a curve) and the other one is physical (as a rate of change)

 

so ..

 

am i trying to find the second y ?

 

 

first let me clear that .. and will continue ..

[mooeypoo]

Where did the square root go?

 

You start out with the big square root, take a derivative of it, and then multiply by the derivative of whatever is inside it (chain rule).

But as you do that, you have another square root. Don't let that scare you, all it means is that you need to multiply the derivative again (chain rule). But as you do THAT derivative, you encounter yet another square root. Again don't fear, just continue the chain rule.

 

It is a chain rule problem. It's just a longer one that requires three chains.

 

Start it out, and I can guide you better once I see how you do the first few steps.

[bimbo36]

Where did the square root go?

 

You start out with the big square root, take a derivative of it, and then multiply by the derivative of whatever is inside it (chain rule).

But as you do that, you have another square root. Don't let that scare you, all it means is that you need to multiply the derivative again (chain rule). But as you do THAT derivative, you encounter yet another square root. Again don't fear, just continue the chain rule.

 

It is a chain rule problem. It's just a longer one that requires three chains.

 

Start it out, and I can guide you better once I see how you do the first few steps.

 

 

the big square root and derivative .. ok .... step (1)

 

X (into)

 

derivative of two more square root of sinx ?

 

 

 

how many terms are we dealing with here ..

 

you know if it was a ploynomial .. there would be two or three terms .. right ?

 

and we usually find the derivative of each term .. yea ?

[Hal.]

....

Edited April 1, 2011 by hal_2011

[bimbo36]

ok .. cool

 

was working my ass off ..lol .. and not reaching anywhere ... exhausted to the mathz ...

 

 

now , since its like a sum .. i should use chain rule ? yes ?

 

 

could you show me how to ?

 

Suppose you want to calculate the derivative of a curve at some particular point. Draw a tangent to the curve at that point. The slope of the tangent gives you the derivative. This is geometrical approach.

 

The algebraical approach is little different. Let y =f(x) which means is a function of x. A very small change in y is denoted by dy and a very small change in x is denoted by dx. The derivative of y means the amount by which y changes if x changes in a very short extent. So mathematically it is given by .

 

 

according to that

 

f(x)=

 

 

so here

 

y = f(x) .. means

 

that ?

 

 

the amount of y changes if x changes in a very short extent .. ?

Edited April 1, 2011 by bimbo36

[Hal.]

My first line ........ dy/dx = 0.5 [ ( sin x + ( ( sin x + ( ( sin x ) ^ 0.5 ) ) ^ 0.5 ) ) ^ - 0.5] [ d/dx ( sin x + ( ( sin x + ( ( sin x ) ^ 0.5 ) ) ^ 0.5 ) ) ]

 

I think that if you can understand this line here you will have no problem finding a solution . Making that solution easier to understand may be difficult but technically you will have an answer that is correct . I am a great fan of brackets !

Edited April 1, 2011 by hal_2011

[bimbo36]

dude i was thinking about putting it in a graph .. one by one

 

in a transparent graph ..

 

one top of another ..

 

 

to help clarify the use of f(x) is saying we have a function that takes x as the input....so f(4) is the same as putting a 4 for every place you see an x on the right hand side of the equality.

 

wtf is this shit then ?

 

 

 

 

this function takes x as input ?

 

 

the whole sugaroot of sin1 + sugar root of sin1 + sugar root of sin 1 ..

 

is different from sugar root of sin1+sugar root of sin1+sugar root of sin 1

 

 

 

 

 

 

 

 

if i had one graph that was like ..

 

f(x)= sugar root of sinx

 

which could take x as input ...

 

 

so f(1) would be like ...

 

f(1)= sugar root of sin1

 

 

 

find dy/dx

 

 

 

silly but if you have .. it would be really nice if you could reply

Edited April 1, 2011 by bimbo36

[mooeypoo]

I think you're getting confused for nothing.

 

Do you know what the derivative of sinx is?

Do you know what the derivative of sqrt(X) is?

 

The first couple of posts were about trying to understand what he question asks. I doubt that they're asking for an infinite series or sum according to what I see from the poster. Do correct me if I'm wrong.

 

Now, to the point. You don't need a graph. You don't need to explain this formula in words. Derivate with chain rule.

 

Just so we know your level (and hence how to helpyou) can you tell us what grade this is for, and if you've done any sort of derivatives and chain rule exercises?

[Cap'n Refsmmat]

This may help:

 

http://www.scienceforums.net/topic/29473-introduction-to-calculus-differentiation/page__view__findpost__p__409619

 

In this case you have several functions inside each other. sin(x) inside a square root, inside a square root with another sin x, etc.

[bimbo36]

I think you're getting confused for nothing.

 

Do you know what the derivative of sinx is?

Do you know what the derivative of sqrt(X) is?

 

The first couple of posts were about trying to understand what he question asks. I doubt that they're asking for an infinite series or sum according to what I see from the poster. Do correct me if I'm wrong.

 

Now, to the point. You don't need a graph. You don't need to explain this formula in words. Derivate with chain rule.

 

Just so we know your level (and hence how to helpyou) can you tell us what grade this is for, and if you've done any sort of derivatives and chain rule exercises?

 

 

if differentiation was about finding slopes ..

 

between points ..

how many points are we dealing with here ?

 

 

 

 

i think the derivative of sinx is cosx ..

 

sqrt of x = x^1/2

 

which is 1/2*x^1/2-1

 

 

isnt it ?

 

 

 

 

 

 

This may help:

 

http://www.sciencefo...post__p__409619

 

In this case you have several functions inside each other. sin(x) inside a square root, inside a square root with another sin x, etc.

 

i am writing that down .. and learning it ..

 

very nice tutorial mr cap

Edited April 2, 2011 by bimbo36

[Fuzzwood]

We arent dealing with points, we are dealing with a function. The point of a derivative of a function is that, when you enter a point from the original function, it gives out the slope of that exact point.

 

You can write that second derivative as [math]\frac{1}{2\sqrt{x}}[/math]

 

Now the next question would be:

 

Do you understand the chain rule? If so, what would be the derivative of sin(-3x) ?

Edited April 2, 2011 by Fuzzwood

[bimbo36]

need a short break and a bit of revisions ..

 

be right back ..

[michel123456]

Nested square roots.

[Hal.]

When you differentiate , what you are trying to do is to get a general expression for the derivative , so that when you then want an individual value , at a specific place , all you have to do is put in the value of x and calculate .

 

The problem with thinking that the slope is from one point to another is that these 2 points ( if we take it as 2 points ) are so close together that the slope is in effect a slope at one point . ( wtf does that mean I hear you say ? An instantaneous slope ! )

 

Just something else about wtf this all means . If we take for an example a car that is moving along at a certain speed , at an instant it is like time has frozen . No time has passed for it to go anywhere or indeed for it to have arrived from anywhere . It still has an instantaneous speed ( derivative of distance with respect to time ) . It still has an instantaneous acceleration ( derivative of speed with respect to time ) .

[mooeypoo]

Guys, I don't mean to be annoying here, but can we try NOT to confuse the poster?

 

This is a derivative question. The poster seems to have the basics semi-clear, so let's concentrate on that and stop posting subjects that make this entire thing overly complicated.

 

bimbo36, your replies to the derivatives are correct:

i think the derivative of sinx is cosx ..

 

sqrt of x = x^1/2

 

which is 1/2*x^1/2-1

Yes.

 

[math]\frac{d}{dx} sin(x) = cos(x) [/math]

 

[math]\frac{d}{dx} \sqrt{x} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}[/math]

Only in your exercise, it's not the square of x, but rather the square of another function, "sinx". What do you do when you have a derivative of a function of x? Chain rule. Take the general derivative and multiply by the derivative of the function itself.

 

So, if I had only this:

[math]\sqrt{sin(x)}[/math]

I will start by considering the square root as its own general function, and multiply by the derivative inside:

 

 

[math]\frac{d}{dx} \sqrt{sin(x)}=\frac{1}{2\sqrt{sinx}} * (\text{(derivative inside:)} cosx)[/math]

So you have the two MAIN derivatives to use. Go read Cap'n's tutorial about the chain rule again, now it's all about organization and multiplying the right chain rules.

 

~mooey

[Cap'n Refsmmat]

Protip: \cos in LaTeX produces

 

[math]\cos x[/math]

 

instead of

 

[math]cos x[/math]

[Hal.]

Should we now ask the original poster if there is enough information here to allow the solving of this problem ?