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 $\sin\theta \cos\theta \tan\theta$ 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 $\frac{dy}{d\theta}$ with respect to $\theta$, 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 $\frac{dx}{d\theta}$ 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).