# Calculus

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Any help would be much appreciated with the steps involved in each problem given. Thank you.

1) Find derivative if y =cot x + sin x

2) Find derivative if y = (3x+5)^8

3) Find derivative if y = x csc x

4) Find derivative if y = x /3x + 1 (3x + 1 is under a square root(/)

5) ( f ◦ g )Œ if f (x) = x^5 + 1 and g (x) = /x at x = 1 (the x is under a square root(/)

Best Regards

• Calculus -

2) use the chain rule where 2 separated 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.

• Calculus -

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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