How do you find the derivative of #(x-1)/(x^2)#?

Question

1503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-21T15:06:17

The derivative with respect to #x# is: #(-x+2)/x^3#

Method 1: Use the quotient Rule:

#d/dx(u/v) = (u'v-uv')/v^2#

#d/dx((x-1)/x^2) = ((1)x^2 - (x-1)*2x)/(x^2)^2#

Now simplify:

#d/dx((x-1)/x^2) = (x^2 - 2x^2+2x)/x^4#

# = (-x^2+2x)/x^4#

# = (x(-x+2))/x^4#

# = (-x+2)/x^3#

Method 2 Rewrite first, then use the power rule.

#(x-1)/(x^2) = x/x^2 - 1/x^2 = x^-1-x^-2#

#d/dx((x-1)/(x^2)) = d/dx(x^-1-x^-2)#

# = -x^-2+2x^-3#

# = -1/x^2+2/x^3#

# = (-x+2)/x^3#