- #1

- 219

- 0

^{2}cos(1/x)

i know how to take the derivative using product and chain rule, but i need to find the derivative using the definition of the derivative. so far i did:

lim [(x+h)

^{2}cos(1/x + h) - x

^{2}cos(1/x)] / h

h ~> 0

[x

^{2}cos(1/x+h) + 2xhcos(1/x+h) + h

^{2}cos(1/x+h) - x

^{2}cos(1/x)] / h

then i took [x

^{2}cos(1/x+h)- x

^{2}cos(1/x)] and i factored out the x

^{2}

x

^{2}[cos(1/x+h) - cos(1/x)]

i used the sum to product formula from trigonometry and i got x

^{2}[-2sin(2x+h / 2x

^{2}+2xh)sin(-h / (2x+h / 2x

^{2}+2xh)]

but from there i'm stuck. i have no idea how to simplify that expression in order to get the h on the bottom of the the entire fraction to cancel out so i can substitute 0 for h. please help.