Find the difference quotient

Question

f(x)= √4x=5

Answer 1

f(x) = sqr4x = 5

2x^1/2 = 5

x^1/2 = 5/2

x = (5/2)^2 = 25/4

My guess is you hit =5 when you meant +5? It's also strange having a sqr sign with 4x, but having it for the 4 only is also strange. If it's just for the 4, then the equation becomes 2x = 5 and x =5/4. You may have copied the problem incorrectly?

but you're asking for the difference quotient that gives you the derivative?

that suggests you meant f(x) = sqr4x + 5 and you want f'(x) = 1/sqrx = (sqrx)x

f(x+h) = sqr(4(x+h)) + 5

difference quotient is [f(x+h)-f(x)]/h

then you take the limit as h goes to zero, and that gives the derivative

sqr(4x+4h) + 5 - sar(4x) + 5 all over h

=[2(sqr(x+h)-2sqrx]/h

multiply numerator and denominator by the conjugate sqr(x+h)+sqrx to get

= 2(x+h-x)/h{sqr(x+h)+sqrx]

=2h/h(sqr(x+h)+sqrx)

=2/sqr(x+h)+sqrx

Now let h = 0

=2/2sqrx

=1/sqrx

which is the same as the expression for the derivative using calculus

If your problems was slightly different, this is still the basic method