Difficult Derivative

[mario99]

\(\displaystyle y(x(\theta))\)

\(\displaystyle \frac{dy}{d\theta} = \frac{dy}{dx} \ \sin\theta \ \cos\theta \ \tan\theta\)

\(\displaystyle \frac{d^2y}{d\theta^2} = \ ?\)

Any help would be appreciated!

 

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[topsquark]

Hint: How would you take the derivative of y = f(x) g(x) h(x)?

-Dan

 

[Dr.Peterson]

\(\displaystyle y(x(\theta))\)

\(\displaystyle \frac{dy}{d\theta} = \frac{dy}{dx} \ \sin\theta \ \cos\theta \ \tan\theta\)

\(\displaystyle \frac{d^2y}{d\theta^2} = \ ?\)

Any help would be appreciated!

First, you are told nothing about y(x), so I presume the answer will look something like what you are given, with unevaluated derivatives of y.

Second, you can simplify [imath]\sin\theta \cos\theta \tan\theta[/imath] before you do anything else.

I would like to see an image of the actual, entire problem, to confirm that nothing has been omitted, as it is an odd problem.

 

[Shiloh]

If y is a function of x which is a function of \(\displaystyle \theta\), then \(\displaystyle \frac{dy}{d\theta}= \frac{dy}{dx}\frac{dx}{d\theta}\).

Here we are told that \(\displaystyle \frac{dy}{d\theta}= \frac{dy}{dx} sin(\theta) cos(\theta) tan(\theta)= \frac{dy}{dx} sin^2(\theta)\).
It follows that \(\displaystyle \frac{dx}{d\theta}= sin^2(\theta)= (1/2)(1- cos(2\theta)\).

\(\displaystyle x= (1/2)\theta- (1/4)sin(2\theta) \) + C

 

[Deleted member 4993]

If y is a function of x which is a function of \(\displaystyle \theta\), then \(\displaystyle \frac{dy}{d\theta}= \frac{dy}{dx}\frac{dx}{d\theta}\).

Here we are told that \(\displaystyle \frac{dy}{d\theta}= \frac{dy}{dx} sin(\theta) cos(\theta) tan(\theta)= \frac{dy}{dx} sin^2(\theta)\).
It follows that \(\displaystyle \frac{dx}{d\theta}= sin^2(\theta)= (1/2)(1- cos(2\theta)\).

\(\displaystyle x= (1/2)\theta- (1/4)sin(2\theta) \) + C

Why are you calculating expression for 'x'?

 

[Shiloh]

So that we can write y and \(\displaystyle \frac{dy}{d\theta}\) as functions of \(\displaystyle \theta\) and find \(\displaystyle \frac{d^2y}{d\theta^2}\).

 

[mario99]

Thank you very much topsquark, Dr.Peterson, Shiloh, and Subhotosh Khan for helping me.

I have to find \(\displaystyle \frac{d^2y}{d\theta^2}\) without simplifying \(\displaystyle \sin\theta \ \cos\theta \ \tan\theta\).

 

[Dr.Peterson]

Thank you very much topsquark, Dr.Peterson, Shiloh, and Subhotosh Khan for helping me.

I have to find \(\displaystyle \frac{d^2y}{d\theta^2}\) without simplifying \(\displaystyle \sin\theta \ \cos\theta \ \tan\theta\).

Once again, please show what you have tried, so we can see where you need help. (See #2) And I'd still very much like to see the original problem (even if it's not in English) to confirm you haven't omitted or misinterpreted something. (See #4)

I hope you see that you need to differentiate [imath]\frac{dy}{d\theta}[/imath] with respect to [imath]\theta[/imath], which will mean using the chain rule and the product rule on the expression you're given. (See #3)

And you do see that you have been told what [imath]\frac{dx}{d\theta}[/imath] is, right?

There's no requirement not to simplify as part of the work; that is always allowed (though not always as helpful as it will be here).