How do you find f'(x) using the definition of a derivative for #f(x)=1/sqrt(x)#?

1 Answer
Jim H
Oct 18, 2015

The crucial step uses #(sqrtx - sqrt(x+h))(sqrtx + sqrt(x+h)) = -h#. If that's not enough of a hint, see below.

Explanation:

#f(x) = 1/sqrtx#

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

# = lim_(hrarr0)(1/sqrt(x+h)-1/sqrtx)/h#

# = lim_(hrarr0)((sqrtx-sqrt(x+h))/(sqrt(x+h)sqrtx))/(h/1)#

# = lim_(hrarr0)((sqrtx-sqrt(x+h)))/((hsqrt(x+h)sqrtx)) #

# = lim_(hrarr0)((sqrtx-sqrt(x+h)))/((hsqrt(x+h)sqrtx)) * ((sqrtx+sqrt(x+h)))/((sqrtx+sqrt(x+h)))#

# = lim_(hrarr0)(x-(x+h))/((hsqrt(x+h)sqrtx)(sqrtx+sqrt(x+h)))#

# = lim_(hrarr0)(-h)/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#

# = lim_(hrarr0)(-1)/(sqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#

# = (-1)/(sqrtxsqrtx(sqrtx+sqrtx))#

# = (-1)/(2xsqrtx)#

Or, if you prefer #(-1)/(2x^(3/2))#

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