question form

#$&*

Mth 271

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

Chain Rule and Product Rule

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I am trying to iron out some of my understanding regarding applicable uses of the chain rule and product rule. I believe i have it down for the most part but i still require insight on a few things and would appreciate any feedback on how i might broaden my understanding.

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For any function that appears to be composed of two functions we use the chain rule. For example y=(3x+1)^2. This could be thought of as the composite of the two functions f(g(x)) where f(g(x)) = x^2 and g(x)=3x+1. We first take the derivative of something with respect to something and multiply it by the derivative of g(x) with respect to x. The something in quotations is basically the derivative of the entire function f(g(x)) which since it is squared would give us 2(g(x)), but i get confused on how this is expressed when writing `dy/`dx. I should be able to recognize this intuitively but the with respects get confusing. should it look like `d(f(g(x)))/`d(f(x)?? The second portion i mentioned already would be expressed `d(g(x))/`d(x).

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Basically the chain rule says that the rate at which a composite of two functions changes is the product of the rates at which the two functions change. We qualify this by specifying that the rates are calculated at the relevant points. Since g acts on x, the rate for the g function is calculated at x. Since f acts on g(x), the rate for the f function is calculated at g(x).

Thus the rate at which f(g) changes when applied at point x is the product of the rate at which g changes at x, which is g ' (x), and the rate at which f changes at g(x), which is f ' (g(x)).

We formulate the chain rule in two different ways. One is consistent with the notation used above:

(f(g(x)) ' = g ' (x) * f ' (g(x)).

The other uses the dy/dx notation, which should not be confused with the f(g(x)) notation. To apply this notation:

Let y = f(z) and z = g(x).

The choice of z is arbitrary. We could use any otherwise unused symbol instead of z. The reason we need a symbol is that our function f is not applied to x, but to g(x). But we don't want to confuse our notation, so we don't use g(x) inside the f function. The reason for this should be clear from the following.

So y = f, a funciton which applies to z, where z = g(x).

So the rate at which g changes with respect to x is dz/dx.

f doesn't act on x but on g(x), which we've denoted as z. So the rate of change of the f function, at its ponit of application, is f ' (z). Since y = f(z), this rate is dy / dz.

So the product of the two relevant rates is

dy / dz * dz / dx.

With this notation we have therefore formed the 'chain' of symbols dy/dz * dz/dx.

This chain isn't a series of algebraic expressions, but in this context it turns out that it can be treated as such, with the dz's 'cancelling' to give us

dy/dz * dz/dx = dy/dx.

Why does this make sense?

If you think about what's been said here, we see that (f(g(x)) ' is just the derivative of y = f(g(x)) with respect to x, so (f(g(x)) ' = dy / dx.

g ' (x) * f ' (g(x)) = dz/dx * dy/dz, which is the same as dy/dz * dz/dx.

So in this notation we've just shown that

dy/dx = dy/dz * dz/dx.

Both notations have their advantages. One advantage of the 'chain' notation is that it can be applied to composites of composites of composites ... for any string of compositions. So for example if y is a function of z, and z a function of u, and u a function of w, with w a function of x and x a function of t, we could write

dy/dt = dy/dz * dz/du * du/dw * dw/dx * dx/dt.

We could equally well decompose the derivative a function of the form

f(g(h(k(t))))

as follows:

(f(g(h(k(t))))) ' = (g(h(k(t)))' * f'(g(h(k(t))))

= (h(k(t))) ' * g ' (h(k(t)) * f ' (g(h(k(t))))

= k ' (t) * h' (k(t)) * g ' (h(k(t)) * f ' (g(h(k(t))))

which probably looks really confusion but isn't bad if you really think about it step by step and remember what everything means.

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The function y=(3x+1)^2 could also be expressed as y=(3x+1)(3x+1) in which case we would apply the product rule. Where we take the derivative of the first function and multiply it by the second function, and add that with the first function times the derivative of the second function. Since 3x+1 cannot be considered the composite of two functions, we only apply the product rule here. I believe these are both potential ways to solve this problem but only one can be used. What are some circumstances in which both are required?

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You can prove some pretty neat stuff using this sort of equivalence. One example is the derivation of derivatives of inverse functions. If f and g are inverses then

(f(g(x)) ' = g ' (x) * f'(g(x)).

But since f and g are inverses f(g(x)) is just x. Thus (f(g(x)) ' = 1.

So g ' (x) = 1 / f ' (g(x)).

Depending of the original function, you now have an expression for the derivative of its inverse function.

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question form

#$&*

Mth 271

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

Product Rule and Chain rule (2

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I submitted a previous inquiry concerning the same thing but I believe this illustrates my difficulties a little better.

I want to think of this problem in terms of the product rule, chain rule, and quotient rule. Starting with the chain rule.

The function that we wish to take the derivative of is f(x)=(5+x^4)^3

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By the chain rule the outer portion (the third power) is differentiated first, then multiplied by the differentiated inner function

=3(5+x^4)^2 *(4x^3)

=12x^3(5+x^4)^2

=12x^3(x^8+10x^4+25)

=12x^11+120x^7+300x^3

So far so good...

Considering the product rule, the function can be expressed

f(x)=(5+x^4)(5+x^4)(5+x^4)

We would proceed as follows:

[`d/`dx(5+x^4)*(5+x^4)*(5+x^4)]+[(5+x^4)*`d/dx(5+x^4)*(5+x^4)]+[(5+x^4)*(5+x^4)*`d/dx(5+x^4)]

=(4x^3)(5+x^4)(5+x^4)+(5+x^4)(4x^3)(5+x^4)+(5+x^4)(5+x^4)(4x^3)

12x^11+120x^7+300x^3

***This is the same answer so both of these were valid ways to solve the problem, however i wish to note that previously i obtained different answers, and this was confusing. However at this point in the question I still wish to show my work and consider the quotient rule as well.

I do not have the quotient rule memorized like the product rule, however I have seen it expressed in terms of the product rule, as follows:

The expression: f(x)=(5+x^4)(5+x^4)(5+x^4) upon further examination this cannot appropriately be solved using the quotient rule. I'll consider the expression as if it were actually:

f(x)=[(5+x^4)/(5+x^4)^2] Still as mentioned above i would solve this as a formation of the product rule by multiplying the numerator by 1/denominator:

=(5+x^4)(1/(x^(8)+10x^(4)+25))

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You could differentiate (5+x^4)/(5+x^4)^2 using the product rule by putting it in the form

(5+x^4) * (5+x^4)^-2

so that the derivative would be

(5 + x^4) / * (5 + x^4)^(-2) + (5 + x^4) * (-2) *(5 + x^4) ' * (5 + x^4)^(-3)

where the derivative of (5 + x^4)^(-2) is (-2) *(5 + x^4) ' * (5 + x^4)^(-3).

I won't work through the rest of the details, but the result would be the same as if you differentiated

(5+x^4)( (x^(8)+10x^(4)+25))^(-1).

More generally, the derivative of f / g could be written as

( f * g^(-1) ) ', where the ^-1 indicates the -1 power as opposed to the inverse function.

The result would be

f ' * (g^(-1)) + f * (g^(-1) ) '

= f ' * g^(-1) + f * (-1) * g ' * g^(-2))

= f ' g^(-1) - f * g ' * g^(-2).

= f' / g - f g ' / g^2

= f ' g / g^2 - f g ' / g^2

= (f ' g - f g ' ) / g^2.

This last form is the usual expression of the quotient rule, and as you can see it can be derived by apply just the product and chain rules to the expression f * g^(-1).

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Of course this would give us a different answer because it is a different function, but the quotient rule can be expressed as the product rule and then solved accordingly. The quotient rule I do not have memorized, but I have found this an effective method of differentiation and would continue to use it.

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Are there circumstances in which either the product rule or quotient rule must and can only be used exclusively? Repeating what was posed in the previous question form, in what circumstances are combinations of chain rule, product rule, and quotient rule ALL used to differentiate. I can imagine chain rule and product or chain and quotient would be appropriate but other possibilities may exist. I have been working diligently to master these concepts which take up a substantial portion of the calculus course. I would appreciate any feedback you can give me, thanks.

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Consider for example the function

x^2 * sqrt( x^2 - 1 ) / (x^3 + x).

The numerator is x^2 sqrt(x^2 - 1), so you would find the derivative of the numerator by applying the product rule to these functions. In the process you would want to find the derivative of sqrt(x^2 - 1), one which you would use the chain rule.

It is easy to find the derivative of the denominator.

By the quotient rule you could then combine the numerator and its derivative with the denominator and its derivative, as f ' g - f g ', to get the numerator of the derivative.

The denominator of the derivative would just be the square of the original denominator.

You could of course apply the product rule to the product

x^2 * sqrt( x^2 - 1 ) ( (x^3 + x)^(-1)

but you would have more simplfying to do at the end. The quotient rule takes care of a fair amount of simplifying, and in the long run it's well worth learning.

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The nice thing about differentiation is that these, along with the rules for differentiating the basic functions, are the only rules you need, and they always work. You can always find the derivative of any function which is written as sums, products, quotients and composites of basic functions whose derivatives are known. It can get messy, and you have to be careful not to lose track of anything, but if you do it right it always works, and with practice you can learn to do it right.

Integration, the inverse operation where you try to figure out what a given function is the derivative of, is much less cooperative. Most of the function that can be written as sums, products, quotients and composites of basic functions whose derivatives are known are not derivatives of such functions, so integration in this sense is usually impossible. When it's possible it's often very tricky. However there are many important functions for which it is possible and not all that tricky, so integration is a very powerful technique for understanding a wide variety of situations.

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