x^4 * (2x+4)^7-Why product rule first and then chain rule?

[xailer]

hello

I understand derivatives, but that understanding ends when I get examples like the ones I posted in this thread.
I'm solving them like a trained monkey, meaning I have no clue of its "internal workings" ( why the procedure actually works).


1.Why do we first apply quotient rule and then chain rule? Why not the other way around?

(3x+4)^2 / (2x+1)

2.Why is derivative of f(x)=e^(2x) equal to 2*e^(2x)? Can you explain the "internal workings" of this formula?

(e^x)' = e^x

f'(x) = e^(2x) * (2x) = 2 * e^(2x) ?

3-Why do we need to apply product rule first and then chain rule? Why not the other way around?

x^4 * (2x+4)^7

thank you

 

[stapel]

1) This is a quotient, so you have to use the Quotient Rule. There is no product, so I'm not sure why you think you would be using the Product Rule at some point...?

2) You have f(x) = e^g(x). Apply the Chain Rule.

3) The Chain Rule applies to a function inside a function. But the "outer" operation here (the first one you get to) is the product, x⁴ times (2x + 4)⁷. Naturally, you apply the Product Rule to the product. Since the second factor of the product involves a nested function (a polynomial sum inside a seventh power), you would then apply the Chain Rule to that factor when processing the Product Rule.

Eliz.

 

[galactus]

Asking such questions shows you'll go further than the average "trained monkey" who goes through the motions just to get it out of their face.

The chain rule allows us to find the derivative of a composition.

Introduce dependent variables:

\(\displaystyle \L\\y=f(g(x))\ and\ u=g(x)\)

then \(\displaystyle \L\\y=f(u)\).

We want the derivatives:

\(\displaystyle \L\\\frac{dy}{du}=f'(u)\) and \(\displaystyle \L\\\frac{du}{dx}=g'(x)\)

We are interested in using known rates of change, dy/du and du/dx, to

find the unknown rate of change dy/dx. Rates of change multiply. i.e. if y

changes 3times the rate of u and u changes 2 times the rate of x, then y

changes 6 times the rate of x.

\(\displaystyle \L\\y=f(g(x))\ and\ u=g(x)\ then\ y=f(u)\)

and \(\displaystyle \L\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\)

Example:

Find \(\displaystyle \L\\\frac{dy}{dx}\ if\ y=4cos(x^{3})\)

Let \(\displaystyle \L\\u=x^{3}\), so that \(\displaystyle \L\\y=4cos(u)\)

Chain rule:

\(\displaystyle \L\\\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{d}{du}[4cos(u)]

\frac{d}{dx}[x^{3}]\)

=\(\displaystyle \L\\(-4sin(u))(3x^{2})=(-4sin(x^{3})(3x^{2})=-12x^{2}sin(x^{3})\)

The formula is easy to remember because if we cancel the du's we get dy\dx.

See how that works?.

 

[xailer]

I understand your first half,but since I haven't yet started learning derivatives of trig functions,I can't comment on last part of your post.

Please,since even I am not shure if this question is totally idiotic or just idiotic, disregard it if you find the former to be true!
I know proof for chain rule, product rule and quotient rule. Proof for product rule, for example, tells me that (f*g)'(x) is not equal to f'(x)+g'(x). Can you think of similar proof that would show me why derivative of function f(x)=x^4 * (2x+4)^7 is equal to :

u = (3x+4)^2
u' = [2*(3x+4)^1]*3        

v = 2x+1
v' = 2

(vu' - uv')/v^2 ?

I'm asking since similary to composition of functions (chain rule) there could be some "internal workings" that make derivative bigger or smaller by some factor (I know there aren't any such "internal workings", so I'm asking for proof that it isn't smaller by some factor), similary to the way we prove with chain rule that derivative of composition of functions is smaller or greater by some factor ?

 

[pka]

If this does not help, please ignore it.
I think that axiler is thinking that the chain rule is quite apart from other rules for differentiation. Of course it is just another tool.
Consider this function: \(\displaystyle f(x) = (x^4) \sqrt {(x^2 - x + 1)}\).
We would say that this is a product. We would use the product rule first and within using the product rule we need the power rule for the square root.

Now for the function: \(\displaystyle g(x) = \sqrt {(x^4) (x^2 - x + 1)}\).
We would say that g is basically a square root.
We would have to use the power rule first for the square root.
Within using the power rule, the chain rule, we need to use the product rule.

Thus it comes down to the order of operations
That determines order in which we use the differentiation tools.