To find the derivative of y = 1/x^2 using first principles, we will start by writing down the definition of the derivative:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
Given that f(x) = 1/x^2, we can substitute into the formula to obtain:
f'(x) = lim(h->0) [(1/(x+h)^2 - 1/x^2) / h]
Next, we will simplify the expression:
f'(x) = lim(h->0) [(1/(x^2 + 2hx + h^2) - 1/x^2) / h]
f'(x) = lim(h->0) [(x^2 - x^2 - 2hx - h^2) / (x^2(x^2 + 2hx + h^2)) / h]
f'(x) = lim(h->0) [(-2hx - h^2) / (x^2(x^2 + 2hx + h^2)) / h]
f'(x) = lim(h->0) [-2hx - h^2 / h(x^2)(x^2 + 2hx + h^2)]
f'(x) = lim(h->0) [-2x - h / x^2(x^2 + 2hx + h^2)]
f'(x) = [-2x - 0 / x^2(x^2)]
f'(x) = -2/x^3
Therefore, the derivative of y = 1/x^2 using first principles is:
y' = -2/x^3
Obtain from the first principle y=1/x^2
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