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.
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
2 answers
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.
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.