- #1

- 87

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I had a question on the chain rule....How would I use the chain rule on a quotient...like if i have 1/(t^4 + 1)^3 , Would I use the quotient rule first, or just start with the chain rule?

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- Thread starter ziddy83
- Start date

- #1

- 87

- 0

I had a question on the chain rule....How would I use the chain rule on a quotient...like if i have 1/(t^4 + 1)^3 , Would I use the quotient rule first, or just start with the chain rule?

- #2

- 28

- 0

(bottom*d of top - top (d of bottom) )/bottom squared

wen u do all that u get (i may be wrong though)

-12t^3/(t^4 +1)^4

- #3

- 28

- 0

like i used the quotient rule there

u can also solve it by using the prduct rule (WHICH I GOT WRONG ON THE TEST ERRR)

- #4

- 87

- 0

cool..thanks man.

- #5

- 290

- 1

Quotient rule:

[tex]f(t) = \frac{p(t)}{q(t)} = \frac{1}{(t^4 + 1)^3}[/tex]

so

[tex]p(t) = 1[/tex]

and

[tex]q(t) = (t^4+1)^3[/tex]

which are both functions of t.

Alternatively, the chain rule:

[tex]f(t) = f(u(t)) = \frac{1}{u^3}[/tex]

where [itex]u(t) = t^4 + 1[/itex]

So we have

[tex]\frac{d}{dt}f(u(t)) = \frac{df}{du}\frac{du}{dt}[/tex]

[tex] = \frac{d}{du}\left( u^{-3} \right) \frac{d}{dt}\left( t^4 + 1 \right) = (-3u^{-4})\cdot (4t^3) = \frac{-12t^3}{(t^4+1)^4}[/tex]

I imagine the chain rule method is a bit faster, and I personally think I'd be more likely to make a silly mistake with the quotient rule, so.

--Justin

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