Question

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

Answer 1

`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`