What is the derivative of #sqrt(x) #?

1 Answer
sente
Jun 8, 2016

#d/dxsqrt(x) = 1/(2sqrt(x))#

Explanation:

In general, the derivative of #x^n# is #d/dxx^n = nx^(n-1)#. From that, we have

#d/dxsqrt(x) = d/dxx^(1/2) = 1/2x^(1/2-1) = 1/2x^(-1/2) = 1/(2x^(1/2))=1/(2sqrt(x))#

We can also use the definition of the derivative:

#f'(x) = lim_(h->0)(f(x+h)-f(x))/h#

Given #f(x) = sqrt(x)#, we can find the derivative as follows:

#d/dxsqrt(x) = lim_(h->0)(sqrt(x+h)-sqrt(x))/h#

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

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

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

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

#=1/(sqrt(x)+sqrt(x))#

#=1/(2sqrt(x))#

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