General solution to the equation step clarification

[lepika2110]

So I was able to do most of the problem on my own but I got stuck between steps 5 and 6. I don't know how the left side gets reduced to d/dtheta(rsec(theta)). If someone can clarify that would be very appreciated

 

[Romsek]

[MATH]\dfrac{d}{d\theta}\left(r \sec (\theta)\right) = \dfrac{dr}{d\theta} \sec(\theta) + r \sec (\theta)\tan(\theta) [/MATH]

 

[lepika2110]

[MATH]\dfrac{d}{d\theta}\left(r \sec (\theta)\right) = \dfrac{dr}{d\theta} \sec(\theta) + r \sec (\theta)\tan(\theta) [/MATH]

Yea, I know that those two are equal. But I wanted to see the process of getting that equality. I dont know how to simply [MATH]\dfrac{dr}{d\theta} \sec(\theta) + r \sec (\theta)\tan(\theta) [/MATH]

 

[Romsek]

Yea, I know that those two are equal. But I wanted to see the process of getting that equality. I dont know how to simply [MATH]\dfrac{dr}{d\theta} \sec(\theta) + r \sec (\theta)\tan(\theta) [/MATH]

you know how to differentiate products?

[MATH]\dfrac{d}{d\theta}(r \sec(\theta)) = \dfrac{d}{d\theta}r \cdot \sec(\theta) + r \cdot \dfrac{d}{d\theta} \sec(\theta) = \\ \dfrac{dr}{d\theta} \sec(\theta) + r \sec(\theta)\tan(\theta)\\~\\ \text{maybe what you're confused about is that $\dfrac{d}{d\theta} \sec(\theta) = \sec(\theta)\tan(\theta)$} [/MATH]

 

[JeffM]

Is there a new notation for partial derivatives?

 

[HallsofIvy]

If, indeed, it is that \(\displaystyle \frac{d}{d\theta} sec(\theta)\)that is bothering you, remember that \(\displaystyle sec(\theta)= \frac{1}{cos(\theta)}\)

Let \(\displaystyle u= cos(\theta)\)so that \(\displaystyle sec(\theta)=\frac{1}{u}\) and then \(\displaystyle \frac{d\frac{1}{u}}{du}= -\frac{1}{u^2}\) and \(\displaystyle \frac{du}{d\theta}= -sin(\theta)\) so that \(\displaystyle \frac{d}{d\theta}sec(\theta)= \frac{d \frac{1}{u}}{du}\frac{du}{d\theta}\)\(\displaystyle = \left(frac{1}{u^2}\right)\left(sin(\theta)\right)\)\(\displaystyle =\frac{sin(\theta)}{cos^2(\theta)}\)\(\displaystyle = \left(\frac{sin(\theta)}{cos(\theta)}\right)\left(\frac{1}{cos(\theta)}\right)= tan(\theta)sec(\theta)\)