Calculus Derivative...Need Help!!!?

Question

I need help on solving these Calculus Derivative...

1. Sin2xCosx

2. (2x-5)/(x^(2)-4)

3. (1)/(x^(2))

4. Sin[ (pi.) / (x^(2) - 5)]

5. 4 sq rt. (x sq rt. (x))

6. f(x) = x^(4)e^(x)

Please show the steps so I can understand how you solve it. Thanks in advance...

Answer 1

1. is a combination of chain rule and product rule.
u=sin2x v=cosx
u'=2cos2x v'=-sinx
u'v+v'u= 2cos2x(cosx)+sinx(sin2x)
2.use difference quotient
u=2x-5 v=x^2-4
u'=2 v'=2x
(u'v-v'u)/v^2 just plug in.
3.same thing...use difference quotient
4.use chain rule
u is (pi/(x^2-5), take derivative of u. then take derivative of sin which becomes cosu. then multiply u' cos u.
5.rewrite sqrt as 4(x(x^1/2))^(1/2) then solve.
6. just use product rule.

sorry if i was frank. a couple of these you might need to do yourself b/c that's how you learn. its not quite that hard... its just pluggin in numbers that's all.

Answer 2

1. Sin2xCosx <-- Use product rule
y' = sin2x(-sinx) +2cosxcosx
= -sinxsin2x + 2cos^2x

2. (2x-5)/(x^(2)-4) <-- Use quotient rule
y' = [2(x^2-4)-2x(2x-5)]/(x^2-4)^2
=[2x^2-8 -4x^2 +10x]/(x^2-4)^2
= (-2x^2 +10x-8)/(x^2-4)^2

3. (1)/(x^(2)) <-- Use quotient rule
y' = -2x/x^4 = -2/x^3

4. Sin[ (pi.) / (x^(2) - 5)]
y' = cos[pi/(x^2-5)][-2pix/(x^2-5)^2]

5. 4(x sq rt. (x))^.5=(4x^1.5)^.5 = 4x^.75
y' = 3/x^.25

6. f(x) = x^(4)e^(x)
y' = x^4 e^x + 4x^3e^x