Use the definition of derivative to compute $dy/dx$, given $y=x+x^{-1}$

Question

I am able to solve the equation using the power rule so I know it is $-x^{-2}+1$.

However the question asks for using the definition of derivative and no matter how I try I cannot seem to get back to the same answer as with the power rule.

Looking for the steps please.

Thanks in advance!

lim{x+h+(1/x+h)-x-(1/x)}/h h->0

= lim{h+(x/(x+h)(x))-((x+h)/x(x+h)}/h h->0

= lim {h+(-h/(x+h)(x))}/h

and this is where I am stuck

Answer 1

• definition $\dfrac{f(x+h)-f(x)}{h}=\dfrac{(x+h)+\frac 1{x+h}-x-\frac 1x}h$
• simplify this by reducing to common denominator and isolate $1$
• you are left with the term $-\frac 1{x(x+h)}$ what is its limit when $h\to 0$.

Answer 2

$$ \frac{x+h+\frac{1}{x+h}-x-\frac{1}{x}}{h} = \frac{x((x+h)^2+1)-(x+h)(x^2+1)}{xh(x+h)} = \frac{x^2-1+h}{x(x+h)} \rightarrow \frac{x^2-1}{x^2} $$

These are the main steps, you can fill it in from here.

Answer 3

$$y^{'}(x_0) = \lim_{x\rightarrow{x_0}}\frac{y(x)-y(x_0)}{x - x_0}= \lim_{x\rightarrow{x_0}}\frac{x+x^{-1}-x_0-x_0^{-1}}{x-x_0} = \lim_{x\rightarrow{x_0}} 1 +\frac{\frac{x_0-x}{xx_0}}{x-x_0}=\lim_{x\rightarrow{x_0}}1-\frac{1}{xx_0}=1-\frac{1}{x_0^{2}}$$ there think we want derivate y at posion $x_0$ , hope this can help you.