Posted by Matt on Sunday, September 30, 2007 at 9:28pm.
2) use the chain rule where 2 seperated parts.
y= (3x+ 5)^8
x= 3x+5
dx= 3
y'= x^8dx
8x^7 dx
put back what x is and get
8*3(3x+5)^7
I just put back the dx as well and moved it to multiply it with the 8
so..
24(3x+5)^7
this is what it should be if I remember how to do this since I'm not looking at my cal book right now.
1) there's an identity for d/dx (cot x) which = csc^2 x. So the answer to 1) is csc^2 x + cos x.
3) use the derivative multiplying rule first *d(second) + second *d(first):
y' = x*(-csc(x)*cot(x)) + csc(x)*1
-- there may be some simplification you can do, but I don't know how much that part matters.
4) remember that the square root of 'something' = (something)^(1/2), so you can rewrite the equation as y = x / (3x+1)^(1/2)
Use the dividing derivative rule of [bottom * d(top) - top * d(bottom)] / bottom^2.
[(3x+1)^(1/2)*1 - x(3x+1)^(1/2)] / {(3x+1)^(1/2)}^2
[(3x+1)^(1/2) - x*(3x+1)^(1/2)] / (3x+1)
5) not sure about the syntax on 5. It may be more clear if you use sqrt(x) or x^(1/2) to denote square roots.
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