Calculus AB problems

<p>what's up guys?
How do you guys solve these problems?
I know these are easy for you guys but not for me!
Please give me a hand.</p>

<li><p>Find dy/dx
<li><p>Find all values of the constants m and b for which the fuction is continuous at x=pi. And Differentiable at x=pi.</p>

<p>sinx, x<pi
mx+b, x is greater than/equal to pi
(y= is in front of two equations)
Many thanx in advance.</p></li>

<p>Heh...We BC students finished that 3 weeks ago :)</p>

<li>I'm assuming you mean y=x^3sin4x instead...
dy/dx = (x^3)(cos4x)(4) + (sin4x)(3x^2)
dy/dx = 4x^3cos4x + 3x^2sin4x</li>

<p>Basically, chain rule on top of product rule</p>

<p>I'm confused about #2...Heh.</p>

<p>I'll give 2 a stab.</p>

<p>Basically you want to find a point at x=pi where sin x would flow right into mx+b. you know that at x=pi, sin x = 0. So an obvious (m,b) pair would be (0,0), since 0*pi + 0 =0. Another acceptable answer is (-3/pi,3). Another, (1/pi, -1). A way of expressing all of these acceptable answers put together would be "(t/pi, -t) where t is any real number." </p>

<p>I think that's what you want, but you might want to have confirmation from some of the other people here.</p>

<p>Seems right. My ending answer was -b/pi = m. Then I guess you pick and choose values. It just seemed odd.</p>