How do you use the limit definition to find the derivative of #2sqrtx-1/(2sqrtx)#?

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Answer 1 · 2017-10-14T18:08:36

Using the lit definition we have:

#d/dx(2sqrtx-1/(2sqrtx)) = lim_(h->0) (2sqrt(x+h) -1/(2sqrt(x+h)) - 2sqrtx+1/(2sqrtx))/h#

#d/dx(2sqrtx-1/(2sqrtx)) = lim_(h->0) (2sqrt(x+h) - 2sqrtx) /h - (1/(2sqrt(x+h))-1/(2sqrtx))/h#

Now rationalize the numerator of the first term:

#(2sqrt(x+h) - 2sqrtx) /h = (2sqrt(x+h) - 2sqrtx) /h xx (sqrt(x+h) + sqrtx)/ (sqrt(x+h) +sqrtx)#

#(2sqrt(x+h) - 2sqrtx) /h = (2(x+h) - 2x) /(h (sqrt(x+h) +sqrtx)#

#(2sqrt(x+h) - 2sqrtx) /h = (2cancelh) /(cancelh (sqrt(x+h) +sqrtx)#

So:

#lim_(h->0) (2sqrt(x+h) - 2sqrtx) /h = 1/sqrtx#

For the second term of the sum:

#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (2sqrtx -2sqrt(x+h))/(hsqrtxsqrt(x+h))#

and in the same way as above:

#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (2x -2(x+h))/(hsqrtxsqrt(x+h)(sqrtx +sqrt(x+h))#

#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (-2cancelh)/(cancelhsqrtxsqrt(x+h)(sqrtx +sqrt(x+h))#

so that:

#lim_(h->0)(1/(2sqrt(x+h))-1/(2sqrtx))/h = -1/(xsqrtx)#

Finally:

#d/dx(2sqrtx-1/(2sqrtx)) = 1/sqrtx+1/(xsqrtx)#