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.
3 is just a constant, not one of the composited functions. We have:
$\displaystyle f = x^4$ and $\displaystyle g = 2-8x$
The derivatives are:
$\displaystyle f' = 4x^3$ and $\displaystyle g' = -8$
The chain rule states that the derivative is $\displaystyle f'[g(x)] * g'(x)$. So the derivative is:
$\displaystyle 3*4(2-8x)^3 * (-8) = -96(2-8x)^3$
Hello, droidus!
Why is the derivative of $\displaystyle 3(2-8x)^4$ 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 $\displaystyle 3x^5$ 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: .$\displaystyle f(x) \:=\:3\cdot x^5$
Using the Product Rule: .$\displaystyle f'(x) \;=\;0\cdot x^5+ 3\cdot5x^4 \;=\; 15x^4$
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: .$\displaystyle f(x)\:=\:\tfrac{1}{2}x^6 $
Re-write it as: .$\displaystyle f(x) \:=\:\frac{x^6}{2}$
Now use the Quotient Rule: .$\displaystyle f'(x) \;=\;\frac{2\cdot 6x^5 - x^6\cdot 0}{2^2} \:=\:\frac{12x^5}{4} \:=\:3x^5$
. . Whee! .Wasn't that fun?
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:
$\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$
$\displaystyle = \lim_{h \to 0} \frac{c(a+h) - ca}{h} = \lim_{h \to 0} \frac{ca + ch - ca}{h}$
$\displaystyle = \lim_{h \to 0} \frac{ch}{h} = \lim_{h \to 0} c = c$
so the derivative of the function f(x) = cx is the number c.
NOW we can use the chain rule.
suppose $\displaystyle g(x) = 3(2 - 8x)^4$. then $\displaystyle g = f \circ h$, where:
$\displaystyle f(x) = 3x$
$\displaystyle h(x) = (2 - 8x)^4$
note that h itself is a composition: $\displaystyle h = k \circ p$, where:
$\displaystyle k(x) = x^4$
$\displaystyle p(x) = 2 - 8x$.
we'll get to h later, first let's use the chain rule on g:
$\displaystyle g'(x) = (f \circ h)'(x) = (f'(h(x)))(h'(x))$
as we saw above, $\displaystyle f'(h(x)) = 3$, no matter what h(x) is (f' is a constant function).
so $\displaystyle g'(x) = 3h'(x)$.
now we use the chain rule on h:
$\displaystyle h'(x) = (k \circ p)'(x) = (k'(p(x)))(p'(x))$
by the power rule, we have $\displaystyle k'(x) = 4x^3$, so $\displaystyle k'(p(x)) = 4(p(x))^3 = 4(2 - 8x)^3$.
finally, we find p'(x) = 0 - 8 = -8.
now we "assemble the pieces":
$\displaystyle g'(x) = 3h'(x) = (3)(4)(2 - 8x)^3(-8) = -96(2 - 8x)^3 = 96(8x - 2)^3$.
Just in case a picture helps...
Ragnarok:
... where (key in spoiler) ...
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