We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
287
3
avatar+564 

 

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

 Oct 14, 2014

Best Answer 

 #3
avatar+102386 
+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
avatar+17776 
+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
avatar+564 
0

Thanks so much Genos!

 Oct 15, 2014
 #3
avatar+102386 
+10
Best Answer

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

19 Online Users

avatar