# Thread: using the chain rule

1. ## using the chain rule

why is the derivative of 3(2-8x)^4 not 0? I thought you take the derivative of the outside (the derivative of 3 is 0), times the inside.

2. ## Re: using the chain rule

3 is just a constant, not one of the composited functions. We have:

$f = x^4$ and $g = 2-8x$

The derivatives are:

$f' = 4x^3$ and $g' = -8$

The chain rule states that the derivative is $f'[g(x)] * g'(x)$. So the derivative is:

$3*4(2-8x)^3 * (-8) = -96(2-8x)^3$

3. ## Re: using the chain rule

Actually, what you are misusing is the product rule, (fg)'= f'g+ fg'. Here, f= 3 and g= (2- 8x)^4. Yes, f'= 0 so we have (3g)'= 0g+ 3g'= 3(4(2- 8x)^3)(-8)= -96(2- 8x)^3

4. ## Re: using the chain rule

Hello, droidus!

Why is the derivative of $3(2-8x)^4$ not 0?
I thought you take the derivative of the outside (the derivative of 3 is 0), times the inside.

WOW! . . . I hope you are on your first day of differentiation.
Otherwise, you need some serious reviewing of the basics.

Would you say that the derivative of $3x^5$ is zero
. . because the derivative of 3 is zero . . . times the "inside"?

Then any expression with a number in front has a derivative of zero?

You are trying (unsuccessfully) to use the Prouct Rule.

We have: . $f(x) \:=\:3\cdot x^5$

Using the Product Rule: . $f'(x) \;=\;0\cdot x^5+ 3\cdot5x^4 \;=\; 15x^4$

Obviously, using the Product Rule when one factor is a constant is a waste of energy

Did you ever use the Quotient Rule just to make it harder?

Example: . $f(x)\:=\:\tfrac{1}{2}x^6$

Re-write it as: . $f(x) \:=\:\frac{x^6}{2}$

Now use the Quotient Rule: . $f'(x) \;=\;\frac{2\cdot 6x^5 - x^6\cdot 0}{2^2} \:=\:\frac{12x^5}{4} \:=\:3x^5$

. . Whee! .Wasn't that fun?

5. ## Re: using the chain rule

when we multiply by a constant, we are employing the function:

f = c*(___), that is:

f(x) = cx.

this is a *different function* than f(x) = c, which has derivative 0.

let's find the derivative of f(x) = cx, straight from the definition:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

$= \lim_{h \to 0} \frac{c(a+h) - ca}{h} = \lim_{h \to 0} \frac{ca + ch - ca}{h}$

$= \lim_{h \to 0} \frac{ch}{h} = \lim_{h \to 0} c = c$

so the derivative of the function f(x) = cx is the number c.

NOW we can use the chain rule.

suppose $g(x) = 3(2 - 8x)^4$. then $g = f \circ h$, where:

$f(x) = 3x$
$h(x) = (2 - 8x)^4$

note that h itself is a composition: $h = k \circ p$, where:

$k(x) = x^4$
$p(x) = 2 - 8x$.

we'll get to h later, first let's use the chain rule on g:

$g'(x) = (f \circ h)'(x) = (f'(h(x)))(h'(x))$

as we saw above, $f'(h(x)) = 3$, no matter what h(x) is (f' is a constant function).

so $g'(x) = 3h'(x)$.

now we use the chain rule on h:

$h'(x) = (k \circ p)'(x) = (k'(p(x)))(p'(x))$

by the power rule, we have $k'(x) = 4x^3$, so $k'(p(x)) = 4(p(x))^3 = 4(2 - 8x)^3$.

finally, we find p'(x) = 0 - 8 = -8.

now we "assemble the pieces":

$g'(x) = 3h'(x) = (3)(4)(2 - 8x)^3(-8) = -96(2 - 8x)^3 = 96(8x - 2)^3$.

6. ## Re: using the chain rule

Just in case a picture helps...

Ragnarok:

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

Halls:

... where (key in spoiler) ...

Spoiler:

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x.

Deveno:

More...

Spoiler:

It's also possible (but of course you'll never usually bother) to boil the power rule down with the product rule, e.g...

This also justifies the multi-factor Product rule - Wikipedia, the free encyclopedia, e.g...

and

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

7. ## Re: using the chain rule

@tom@ballooncalculus

i like how you turn the commutativity of addition (of real-valued functions) into equivalence of "diagrams" in your picture of the product rule. (to turn the first into the second, we "untwist it").