Pokémon TCG: Sword and Shield—Brilliant Stars

help me with math =/

B

BlazeUp

New Member
i have no idea how to do this..

If f'(x) = x^3 and u = e^x, show that d/dx(f(u)) = e^4x

i dont evne know where to begin...
 
Yay calculus.

First (and you probably know this), f'(x) is shorthand for df/dx
so if f'(x) = x^3
then f(x) = (1/4)(x^4) (anti-derivative of x^3)

So f(u) = f(e^x)
= (1/4)((e^x)^4)
= (1/4)(e^4x)​

d/dx((1/4)(e^4x)) = (1/4) d/dx(e^4x) (Constant rule: If k is constant then d/dx(kx) = k(d/dx(x)))

Now apply the chain rule to the new derivative:
If h(x) = f(g(x))
then h'(x) = f'(g(x)) g'(x)​
Letting
f(x) = e^x​
and
g(x) = 4x​
Now derive:

f(x) = e^x => f'(x)= e^x (e^x is its own derivative)

g(x) = 4x => g'(x) = (4)(1x^0) = 4

So h'(x) = (e^(4x))(4)
= 4e^4x​

So back to d/dx((1/4)(e^4x))
= (1/4) d/dx(e^4x)
= (1/4) (4e^4x)
= (1/4) (4) (e^4x)
= e^4x​

I repeat: yay calculus.

And I eagerly await a bulletin board that can handle MathML.

Also, I take the title "Pokemon Professor" quite literally.
 
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lol weird. I've just been doing this exact same stuff this week. You guys use slightly different notations to us but its all the same in the end. math <3
 
Sheesh that's complicated. Why not just say d/dx(f(u)) = f'(u)u' = [(e^x)^3][e^x] = (e^3x)(e^x) = e^4x ?

Also, technically, you should probably have a constant at the end of your integral. Even though you're just going to take the derivative immediately.
 
I just posted here to see whats going on in this thread.By the way till high standard i had not heard about this chain rule.Which standard do u read?
 
pat460 said:
No offense or anything, but why did you ask for math help on pokegym?
Click to expand...

Because it was a Random "Topic!"

On top of that, there are some VERY intelligent people on Pokegym, that have expertises is different areas...The above happened to be in "Math.":thumb:
 
ZAKtheGeek said:
Sheesh that's complicated. Why not just say d/dx(f(u)) = f'(u)u' = [(e^x)^3][e^x] = (e^3x)(e^x) = e^4x ?

Also, technically, you should probably have a constant at the end of your integral. Even though you're just going to take the derivative immediately.
Click to expand...

The first explanation is good because it is far easier to use if the problem had more complicated derivatives... but yes, forgetting constants of integration is a big no-no :rolleyes:

They give the same answer in the end - use the one you like best.
 
dogma said:
The first explanation is good because it is far easier to use if the problem had more complicated derivatives... but yes, forgetting constants of indefinite integration is a big no-no :rolleyes:

They give the same answer in the end - use the one you like best.
Click to expand...

fixed
 
I edited the quote below to fix your formatting to indeces instead of ^s:

Chairman Kaga said:
Yay calculus.

First (and you probably know this), f'(x) is shorthand for df/dx
so if f'(x) = x[sup]3[/sup]
then f(x) = (1/4)(x[sup]4[/sup]) (anti-derivative of x[sup]3[/sup])

So f(u) = f(e[sup]x[/sup])
= (1/4)((e[sup]x[/sup])[sup]4[/sup])
= (1/4)(e[sup]4x[/sup])​

d/dx((1/4)(e[sup]4x[/sup])) = (1/4) d/dx(e[sup]4x[/sup]) (Constant rule: If k is constant then d/dx(kx) = k(d/dx(x)))

Now apply the chain rule to the new derivative:
If h(x) = f(g(x))
then h'(x) = f'(g(x)) g'(x)​
Letting
f(x) = e[sup]x[/sup]​
and
g(x) = 4x​
Now derive:

f(x) = e[sup]x[/sup] => f'(x)= e[sup]x[/sup] (e[sup]x[/sup] is its own derivative)

g(x) = 4x => g'(x) = (4)(1x[sup]0[/sup]) = 4

So h'(x) = (e[sup]4x[/sup])(4)
= 4e[sup]4x[/sup]​

So back to d/dx((1/4)([sup]4x[/sup]))
= (1/4) d/dx(e[sup]4x[/sup])
= (1/4) (4e[sup]4x[/sup])
= (1/4) (4) (e[sup]4x[/sup])
= e[sup]4x[/sup]​

I repeat: yay calculus.

And I eagerly await a bulletin board that can handle MathML.

Also, I take the title "Pokemon Professor" quite literally.
Click to expand...
 
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