Derivative of y=cos(a3+x3)What is the derivative of y=cos(a3+x3)?

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The derivative of y=cos(a^3+x^3) is calculated using the chain rule. The derivative is found to be dy/dx = -3sin(a^3+x^3)(a^2+x^2). The original poster realizes that they incorrectly included a term for a^3, which is a constant, thus its derivative should be zero. They also acknowledge a misunderstanding regarding the use of SIZE commands in the forum. Clear guidance is provided on how to format equations correctly for better readability.
physics604
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Homework Statement



Find the derivative of y=cos(a3+x3)

Homework Equations



Chain rule

The Attempt at a Solution



y=cosu

\frac{dy}{du} = -sinu

u=a3+x3

\frac{du}{dx} = 3a2+3x2


\frac{dy}{dx} = -3sin(a3+x3)(a2+x2).

The answer is supposed to be -3x2sin(a3+x3). What did I do wrong?

Any help is much appreciated.
 
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physics604 said:

Homework Statement



Find the derivative of y=cos(a3+x3)

Homework Equations



Chain rule

The Attempt at a Solution



y=cosu

\frac{dy}{du} = -sinu

u=a3+x3

\frac{du}{dx} = 3a2+3x2



##a## is constant so the derivative of ##a^3## is ##0##.

Please do not insert SIZE commands in your posts. It makes them very hard to follow when quoting them, aside from violating forum policy.
 
Okay, thanks!

I was thinking maybe that was my error, but I wasn't 100% sure.

And I didn't know about the SIZE commands rule, so I'll stay away from them from now on. I just thought they'd be easier to read since I don't know how to make the equations like yours.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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