# Thread: Differentiate and find the square of dy/dx

1. ## Differentiate and find the square of dy/dx

Question:

If $\displaystyle x = cos \theta + \theta sin \theta$ , $\displaystyle y = sin \theta - \theta cos \theta$ , then show that

$\displaystyle \frac{dy}{dx} = tan \theta$

Also find $\displaystyle \frac{d^2 y}{dx^2}$

2. Originally Posted by zorro
Question:

If $\displaystyle x = cos \theta + \theta sin \theta$ , $\displaystyle y = sin \theta - \theta cos \theta$ , then show that

$\displaystyle \frac{dy}{dx} = tan \theta$

Also find $\displaystyle \frac{d^2 y}{dx^2}$
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (which I'm sure will be somewhere in yur class notes or textbook).

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}$

If you need more help, please show all your working and state specifically where you are stuck.

3. "yur"?

4. ## Is this correct?

Originally Posted by mr fantastic
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (which I'm sure will be somewhere in yur class notes or textbook).

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}$

If you need more help, please show all your working and state specifically where you are stuck.

I am getting the $\displaystyle \frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}$

$\displaystyle \frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta$

$\displaystyle \frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}$

therefore

$\displaystyle \frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}$

Is this Right!!!

5. Originally Posted by zorro
I am getting the $\displaystyle \frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}$

$\displaystyle \frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta$

$\displaystyle \frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}$

therefore

$\displaystyle \frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}$

Is this Right!!!
Yes, and if you recall that $\displaystyle \frac{1}{\cos \theta} = \sec \theta$ then you can simplify your answer a little bit.

6. ## So after simplifying

Originally Posted by mr fantastic
Yes, and if you recall that $\displaystyle \frac{1}{\cos \theta} = \sec \theta$ then you can simplify your answer a little bit.
= $\displaystyle sec^2 \theta . \frac{1}{\theta cos \theta}$
= $\displaystyle \frac{sec^3 \theta}{ \theta}$

Is this right now???

7. Originally Posted by zorro
= $\displaystyle sec^2 \theta . \frac{1}{\theta cos \theta}$
= $\displaystyle \frac{sec^3 \theta}{ \theta}$

Is this right now???
Correct.

8. By the way, Zorro, $\displaystyle \frac{d^2y}{dx^2}$ is NOT "the square of dy/dx"!

9. ## Thanks u every one for helping me

Thank u Mr fantastic for ur help