Solving the Chain Rule: A Visual Guide

[QuarkCharmer]

Homework Statement

\frac{d}{dx}(x+(x+sin^2(x))^3)^4

Homework Equations

Calc up to Chain Rule.

The Attempt at a Solution

Using product and chain rule I got:
\frac{dy}{dx}=4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2)(1+\frac{d}{dx}sin^2(x))

Then I calculated the derivative of sin^2(x):
\frac{d}{dx}sin^2(x)=2sin(x)cos(x)

and put that into the derivative to get:

y'=4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2)(1+2sin(x)cos(x))

Do I further simplify this? It does not seem obvious to me. What should I be paying attention to next?
Here is another problem I worked out. I think it's correct, assuming I am using the chain-rule correctly?
[PLAIN]http://img824.imageshack.us/img824/2691/imag0025ed.jpg
It simplified down to something sort of pretty, but that other one looks horrible!

 

[itheo92]

That is correct. I think you can't simplify it further. But at entry level, if you're afraid of using chain rule for really lengthy functions, like this, you can substitute a suitable part of it. ie.

\frac{d}{dx}(x+(x+sin^2(x))^3)^4
let,
(x+sin^2(x)) = z
so,
\frac{d}{dx}(x+(x+sin^2(x))^3)^4 = \frac{d}{dx}(x+z^3)^4

Now, do differentiate in the same way. The final answer will be same, but it may decrease the chance to make a mistake.

 

[QuarkCharmer]

That is basically what I was doing. Only I did it in my head.

 

[tiny-tim]

That is basically what I was doing. Only I did it in my head.

Hi QuarkCharmer!

I think you have a bracket in the wrong place, but otherwise it's fine.

In your second one, if you're wondering why it's so neat, that's because you could rewrite the bottom as (√2)cos(πx - π/4) …

then the top becomes cossin - sincos = sin((πx - π/4) - πx) = sin(-π/4) = -1/√2.

 

[QuarkCharmer]

Hi QuarkCharmer!

I think you have a bracket in the wrong place, but otherwise it's fine.

In your second one, if you're wondering why it's so neat, that's because you could rewrite the bottom as (√2)cos(πx - π/4) …

then the top becomes cossin - sincos = sin((πx - π/4) - πx) = sin(-π/4) = -1/√2.

I see what you did there, but unless it's really obvious, those phase shifting identities would be going a bit too far I would think?

 

[tiny-tim]

Hi QuarkCharmer!

… phase shifting identities …

You make it sound like something out of Star Trek!

Surely it's always simpler to deal with one cos rather than a sum of two?

(You might like to try differentiating a general cos(x+A)/cosx, = cosA + sinAtanx )

 

[QuarkCharmer]

Hi QuarkCharmer!


You make it sound like something out of Star Trek!

Surely it's always simpler to deal with one cos rather than a sum of two?

(You might like to try differentiating a general cos(x+A)/cosx, = cosA + sinAtanx )

Hey, I didn't make up the nomenclature! I learned about alternating current before I knew what trigonometry was, so I just think of it that way.

I'll give that a shot, but do you mean to prove the equality, or differentiate one of the sides?

 

[tiny-tim]

I hope the identity is obvious!

I meant differentiate one of the sides. ​