Asked by Anonymous
differentiation form first principles y=1/x^2
Answers
Answered by
drwls
form or from first principles?
The derivative of y = x^-2 is
dy/dx = -2 x^-3
For a 'first principles' derivation, you will have to use the definition of the derivative and calculate some limits.
I recommend that you perform that exercise yourself.
dy/dx = lim (1/d)[1/(x+d)^2 - 1/d^2]
.......d -> 0
The derivative of y = x^-2 is
dy/dx = -2 x^-3
For a 'first principles' derivation, you will have to use the definition of the derivative and calculate some limits.
I recommend that you perform that exercise yourself.
dy/dx = lim (1/d)[1/(x+d)^2 - 1/d^2]
.......d -> 0
Answered by
Anonymous
i get so far then i get stuck i think im doing something wrong
f(x+h)-f(x) = 1/(x+h)^2 - 1/x^2
= x^2 - x^2 -2xh -h^2 above divided by(x^2+2xh+h^2)(x^2)
= -2xh -h^2/x^4+2x^3h+h^2x^2
then i know i need to divide by h but i cant seem to get to what i want
f(x+h)-f(x) = 1/(x+h)^2 - 1/x^2
= x^2 - x^2 -2xh -h^2 above divided by(x^2+2xh+h^2)(x^2)
= -2xh -h^2/x^4+2x^3h+h^2x^2
then i know i need to divide by h but i cant seem to get to what i want
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