Find the derivative of $x^{- 1}$ $x^(-1)$ from the definition of derivative?

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Answer 1 · 2017-02-17T17:00:22

$\frac{d}{d x} x^{- 1} = - \frac{1}{x^{2}}$ $d/(dx) x^(-1)=-1/x^2$

From the definition of derivative of a function $f (x)$ $f(x)$ ,

$\frac{d}{d x} f (x) = L t_{h \to 0} \frac{f (x + h) - f (x)}{h}$ $d/(dx) f(x)=Lt_(h->0)(f(x+h)-f(x))/h$

As $f (x) = x^{- 1} = \frac{1}{x}$ $f(x)=x^(-1)=1/x$

$\frac{d}{d x} f (x) = L t_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$ $d/(dx) f(x)=Lt_(h->0)(1/(x+h)-1/x)/h$

= $L t_{h \to 0} \frac{\frac{x - x - h}{x^{2} + h x}}{h}$ $Lt_(h->0)((x-x-h)/(x^2+hx))/h$

= $L t_{h \to 0} \frac{- h}{h (x^{2} + h x)}$ $Lt_(h->0)(-h)/(h(x^2+hx))$

= $L t_{h \to 0} \frac{- 1}{x^{2} + h x}$ $Lt_(h->0)(-1)/(x^2+hx)$

= $- \frac{1}{x^{2}}$ $-1/x^2$

Find the derivative of x^(-1)x−1x^(-1) from the definition of derivative?