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# Help with derivatives?

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3 I know how to do part A. The parts after however, are giving me some trouble :/

Oct 14, 2014

#3
+10

Gotta' slightly disagree with (f)...we have......

w = re^(-r)   ...... " r " must be a variable or else w' would just be "0"

So we have, using the product rule

w' = e^(-r) + (-1)re^(-r) =   e^(-r)* [ 1 - r]   Oct 15, 2014

#1
+5

Part b:  y = (1/3)x^3 + (1/2)x^2 + (1/4)x

y'  =  3(1/3)x^2 + 2(1/2)x^1 + 1(1/4)x^0

y'  =  x^2 + x + 1/4             The derivative of a constant is zero.

y''  = 2x + 1

c)  s = -2t^-1 + 4t^-2                    (Write as negative exponents; it's easier.)

x' = -1(-2t^-2) + -2(4t^-3)

x' = 2t^-2 - 8t^-3

x'' = -4t^-3 + 24t^-4

d)  r = 12θ^-1 - 4θ^-3 + θ^-4

r' = -12θ^-2 + 12θ^-4 - 4θ^-5

r'' = 24θ^-3 - 48θ^-5 + 20θ^-6

e)  r  = 2θ^(-1/2) + 2θ^(1/2)              Easier to work if negative, fractional exponents.

r'  = -θ^(-3/2) + θ^(-1/2)               Subtract 1 from each exponent

r'' = (3/2)θ^(-5/2) - (1/2)θ^(-3/2)

f)  w = re^(-r)

w' = -re^(-r)                           (Do you have exponentials?)

g)  y = x^(-3/5) + π^(3/2)            π^(3/2) is a constant, so its derivative is zero

y' = (-3/5)x^(-8/5)                  Subtract 1 from the exponent

Oct 14, 2014
#2
0

Thanks so much Genos!

Oct 15, 2014
#3
+10

Gotta' slightly disagree with (f)...we have......

w = re^(-r)   ...... " r " must be a variable or else w' would just be "0"

So we have, using the product rule

w' = e^(-r) + (-1)re^(-r) =   e^(-r)* [ 1 - r]   CPhill Oct 15, 2014