Let f(x) = 1/x^2 on the interval (0, infinity). Using the definition of the deri

Question

Let f(x) = 1/x^2 on the interval (0, infinity). Using the definition of the derivative, find f'(c) for c>0. **definition of derivative: Let f: A --> R be defined on an interval A. Let c be an element of A. The derivative of f(x) at c is defined by f'(c) = lim [f(x) - f(c)] / (x-c) as x approaches c if the limit exists.

Explanation / Answer

f(x) = 1/x^2 by definition, f'(x) = limx->c[f(x) - f(c)]/(x-c) f'(x) = limx->c[1/x^2 - 1/c^2]/(x-c) f'(x) = limx->c[(c^2 - x^2)/x^2c^2(x-c) f'(x) = limx->c(c+x)(c-x)/-x^2c^2(c-x) f'(x) = limx->c[-(c+x)/x^2c^2] putting limits f'(c) = -2c/c^4 f'(c) = -2/c^3 hence f'(x) = -2/x^3

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