Find the derivative of f(x)= x^3/2 using the limit definition ?

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Answer 1 · 2018-05-17T15:16:19

#d/dx (x^(3/2)) =(3sqrtx )/2#

Using the definition of derivative:

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

so, for #f(x) = x^(3/2) = xsqrtx#

#d/dx (x^(3/2)) = lim_(h->0) ((x+h)sqrt(x+h) -xsqrtx)/h#

#d/dx (x^(3/2)) = lim_(h->0) (xsqrt(x+h)+hsqrt(x+h) -xsqrtx)/h#

#d/dx (x^(3/2)) = lim_(h->0) (sqrt(x+h)+ x(sqrt(x+h)-sqrtx)/h)#

#d/dx (x^(3/2)) = lim_(h->0) sqrt(x+h)+ lim_(h->0) x(sqrt(x+h)-sqrtx)/h#

#d/dx (x^(3/2)) =sqrtx+ x lim_(h->0) (sqrt(x+h)-sqrtx)/h#

Multiply and divide the fraction by #(sqrt(x+h)+sqrtx)# and use the algebraic identity: #(a+b)(a-b) = a^2+b^2#:

#d/dx (x^(3/2)) =sqrtx+ x lim_(h->0) ((sqrt(x+h)-sqrtx)(sqrt(x+h)+sqrtx))/(h(sqrt(x+h)+sqrtx))#

#d/dx (x^(3/2)) =sqrtx+ x lim_(h->0) (x+h-x)/(h(sqrt(x+h)+sqrtx))#

#d/dx (x^(3/2)) =sqrtx+ x lim_(h->0) h/(h(sqrt(x+h)+sqrtx))#

#d/dx (x^(3/2)) =sqrtx+ x lim_(h->0) 1/(sqrt(x+h)+sqrtx)#

#d/dx (x^(3/2)) =sqrtx+ x /(2sqrtx)#

#d/dx (x^(3/2)) =sqrtx+ sqrtx /2#

#d/dx (x^(3/2)) =(3sqrtx )/2#