Use the limit definition of a derivative to differentiate an equation:

Question

Use the limit definition of a derivative to differentiate y = √(2x + 3).

I got f'(x) = limh->0(1/(√(2x+3)) is this correct?

My steps were: (using the limit definition of a derivative formula)

f'(x) = limh->0((√(2(x+h) + 3) - √(2x+3))/h

Substitute in 0 for h:

f'(x) = limh->0((√(2(x+0) + 3) - √(2x+3))/0

Answer 1

f(x+h) = sqrt( 2(x+h)+3)

= sqrt(2x+2h+3)

The NUMERATOR of the difference quotient is:

f(x+h) - f(x) = sqrt(2x+2h+3) - sqrt(2x+3)

the denominator of course is just h...

Division by zero not allowed so must rationalize

the numerator, and hope the h cancels out...

the conjugate is sqrt(2x+2h+3) + sqrt(2x+3)

Multiplies the top and bottom by this conjugate...

the difference quotient becomes:

(2x+2h+3) - (2x+3) all over {h [sqrt(2x+2h+3) +sqrt(2x+3)]}

However the numerator simplifies to:

2x+2h+3 -2x - 3 = 2h

the h cancels out as anticipates...

the final difference quotient is:

2 over [sqrt(2x+2h+3)+sqrt(2x+3)] =

as h-->0, the limit is 2 over { 2 [ sqrt(2x+3)] }

1/ sqrt(2x+3)

---------------------------------------------------------

so then as to check the answer by power rule/chain rule:

Dx[ sqrt(2x+3)] = Dx (2x+3)^(1/2)

(1/2) (2x+3)^(-1/2) (2) =

1/sqrt(2x+3)