How do you find f'(x) using the limit definition given #sqrt(2x) - x^3 #?

1 Answer
Dec 8, 2016

# f'(x) = sqrt(2)/(2sqrt(x) ) -3x^2#

Explanation:

The definition of the derivative of #y=f(x)# is

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

So Let # f(x) = sqrt(2)sqrt(x)-x^3 = sqrt(2)sqrt(x)-x^3# then;

# \ \ \ \ \ f(x+h) = sqrt(2)sqrt(x+h)-(x+h)^3 #
# :. f(x+h) = sqrt(2)sqrt(x+h)-(x^3+3x^2h+3xh^+h^3) #
# :. f(x+h) = sqrt(2)sqrt(x+h)-x^3-3x^2h-3xh^-h^3) #

And so the derivative of #y=f(x)# is given by:
# \ \ \ \ \ f'(x) = lim_(h rarr 0) ( (sqrt(2)sqrt(x+h)-x^3-3x^2h-3xh^2-h^3) - (sqrt(2)sqrt(x)-x^3) ) / h #

# :. f'(x) = lim_(h rarr 0) ( sqrt(2)sqrt(x+h)-x^3-3x^2h-3xh^2-h^3 - sqrt(2)sqrt(x)+x^3 ) / h #

# :. f'(x) = lim_(h rarr 0) ( sqrt(2)sqrt(x+h)- sqrt(2)sqrt(x)-3x^2h-3xh^2-h^3 ) / h #

# :. f'(x) = lim_(h rarr 0) { (sqrt(2)sqrt(x+h)- sqrt(2)sqrt(x) )/h -3x^2-3xh-h^2 }#

# :. f'(x) = lim_(h rarr 0) (sqrt(2)/h(sqrt(x+h)- sqrt(x) )) -3x^2#

# :. f'(x) = -3x^2 + lim_(h rarr 0) sqrt(2)/h(sqrt(x+h)- sqrt(x) )*(sqrt(x+h)+ sqrt(x) )/(sqrt(x+h)+ sqrt(x) ) #

# :. f'(x) = -3x^2 + lim_(h rarr 0) sqrt(2)/h((x+h)-(x))/(sqrt(x+h)+ sqrt(x) ) #

# :. f'(x) = -3x^2 + lim_(h rarr 0) sqrt(2)/h(h)/(sqrt(x+h)+ sqrt(x) ) #
# :. f'(x) = -3x^2 + lim_(h rarr 0) sqrt(2)/(sqrt(x+h)+ sqrt(x) ) #
# :. f'(x) = -3x^2 + sqrt(2)/(sqrt(x)+ sqrt(x) ) #
# :. f'(x) = sqrt(2)/(2sqrt(x) ) -3x^2#