How do you take the derivative of -(t + 1)sin(\frac{t^2}{2})?

[firefly_1]

[SOLVED] Refresher on Derivatives

I feel rather dumb asking this but I could use a quick refresher on some parts of derivatives since for the past months I have been solely working on integrals and known cross sections so suddenly switching back has left my brain reeling.

The main question I need explained is how do you take a derivative of $$-(t + 1)sin(\frac{t^2}{2})$$

I remember a trick my AP teacher told me of 1D2 + 2D1 meaning take the first part muliplied by the derivative of the second plus the second multiplied by the derivative of the first. The problem is, I don't remember how to take the derivatve of (t^2)/2

 

[ZioX]

Power rule. (t^n)'=nt^(n-1). Remember that constants slide out as well.

 

[neutrino]

You need to use a combination of the product rule and chain rule to differentiate the whole expression. But for t^2/2, use the power rule.

For a quick refresher of the rules, you can have a look at this thread at the tutorial
https://www.physicsforums.com/showthread.php?t=139690

 

[murshid_islam]

$$\frac{d}{dt}\left[-(t + 1)sin(\frac{t^2}{2})\right]$$

$$= -\left[(t+1)\frac{d}{dt}\left(\sin \frac{t^2}{2}\right) + \left(\sin \frac{t^2}{2}\right)\frac{d}{dt}(t+1)\right]$$

$$= -\left[(t+1)\cos \left(\frac{t^2}{2}\right)\frac{d}{dt}\left(\frac{t^2}{2}\right) + \sin \frac{t^2}{2}\right]$$

$$= -\left[(t+1)\cos \left(\frac{t^2}{2}\right)\cdot t + \sin \frac{t^2}{2}\right]$$