Find and simplify the difference quotient (f(x+h) - f(x))/h for f(x) = sqrt(5x + 1).

Step-by-step solution

1. 1

   Write out the quotient. With $f(x+h) = \sqrt{5(x+h)+1} = \sqrt{5x+5h+1}$,

   $\frac{\sqrt{5x+5h+1} - \sqrt{5x+1}}{h}$

   Nothing cancels yet: the $h$ is buried inside a radical, so it cannot be factored out of the numerator directly.

2. 2

   Multiply by the conjugate of the numerator. This is the standard move for a difference of square roots:

   $\frac{\sqrt{5x+5h+1} - \sqrt{5x+1}}{h} \cdot \frac{\sqrt{5x+5h+1} + \sqrt{5x+1}}{\sqrt{5x+5h+1} + \sqrt{5x+1}}$

   Multiplying by $1$ in this form changes nothing but moves the radicals to the denominator.

3. 3

   Use (A − B)(A + B) = A² − B² on the numerator. The square roots vanish:

   $\left(5x+5h+1\right) - \left(5x+1\right) = 5h$

4. 4

   Cancel the h.

   $\frac{5h}{h\left(\sqrt{5x+5h+1} + \sqrt{5x+1}\right)} = \frac{5}{\sqrt{5x+5h+1} + \sqrt{5x+1}}$

5. 5

   Check numerically and take the limit. At $x = 1.3$, $h = 0.07$: the direct quotient is $0.9024623$ and the formula gives $\tfrac{5}{\sqrt{7.85}+\sqrt{7.5}} = 0.9024623$. As $h \to 0$ both radicals become $\sqrt{5x+1}$, giving $\tfrac{5}{2\sqrt{5x+1}}$ — the derivative of $\sqrt{5x+1}$.