Square roots. The 5-step dance works, but you can't subtract $\sqrt{1+h} - \sqrt{1}$ and just "expand." We need a new trick — multiplying by the conjugate — to get a form where the $h$ is visible and cancelable.
The function $\sqrt{x}$ grows slower than $x$ does (and slower than $2x$, and slower than $10x$). So the slope at $x = 1$ should be small. Specifically, since $\sqrt{x}$ is concave down everywhere, the slope is shrinking as $x$ grows — at $x = 1$ the slope should be $1/2$ (or $0.5$ rounded). That's a useful sanity check before you do the algebra.
The "new trick" the LMS is hinting at is the conjugate. We multiply by $\dfrac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1}$ — which equals $1$, so it doesn't change the value. But it does change the form, in exactly the way we need: the $(\sqrt{} - \sqrt{})(\sqrt{} + \sqrt{}) = (\text{something})^2 - (\text{something else})^2$ pattern is a difference of squares, and the $h$ hides inside one of the squared terms.
The identity we use is the difference of squares:
$$(A - B)(A + B) \;=\; A^2 - B^2$$With $A = \sqrt{1+h}$ and $B = \sqrt{1} = 1$:
$$(\sqrt{1+h} - 1)(\sqrt{1+h} + 1) \;=\; (\sqrt{1+h})^2 - 1^2 \;=\; (1+h) - 1 \;=\; h$$That's how the $h$ shows up — the conjugate turns a difference of square roots into a single $h$. That's the whole magic of the trick.
Given $f(x) = \sqrt{x}$ and $a = 1$. We compute $\displaystyle f'(1) = \lim_{h \to 0} \dfrac{f(1+h) - f(1)}{h}$ in five steps.
Step 1 — Write the difference quotient (and multiply by the conjugate).
$$f'(1) \;=\; \lim_{h \to 0} \frac{\sqrt{1+h} - \sqrt{1}}{h} \cdot \frac{\sqrt{1+h} + \sqrt{1}}{\sqrt{1+h} + \sqrt{1}}$$The $\dfrac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1}$ factor equals $1$, so the limit is unchanged. We're buying a friendlier numerator.
Step 2 — FOIL the numerator.
$$(\sqrt{1+h} - 1)(\sqrt{1+h} + 1) \;=\; (\sqrt{1+h})^2 + \sqrt{1+h} - \sqrt{1+h} - 1$$The two "middle" $\sqrt{1+h}$ terms cancel (one is $+$, the other is $-$). What's left:
$$=\; (\sqrt{1+h})^2 - 1 \;=\; (1+h) - 1 \;=\; h$$The $h$ revealed itself — that's exactly why the conjugate trick works.
Step 3 — Rewrite the whole quotient.
$$f'(1) \;=\; \lim_{h \to 0} \frac{h}{h \cdot (\sqrt{1+h} + 1)} \;=\; \lim_{h \to 0} \frac{h}{h(\sqrt{1+h} + 1)}$$Since $\sqrt{1} = 1$, the conjugate $\sqrt{1+h} + \sqrt{1}$ in the denominator becomes $\sqrt{1+h} + 1$, and it picks up a factor of $h$ from the original denominator.
Step 4 — Cancel $h$.
$$=\; \lim_{h \to 0} \frac{\cancel{h}}{\cancel{h}(\sqrt{1+h} + 1)} \;=\; \lim_{h \to 0} \frac{1}{\sqrt{1+h} + 1}$$Step 5 — Take the limit as $h \to 0$.
$$f'(1) \;=\; \frac{1}{\sqrt{1+0} + 1} \;=\; \frac{1}{1 + 1} \;=\; \frac{1}{2} \;=\; 0.5$$Final answer:
The reason the conjugate trick is needed is that $\sqrt{1+h} - 1$ doesn't have a "natural" $h$ in it — you can't just expand it like $(x+h)^2$. So the numerator is a black box with an $h$ trapped inside a square root, and there's no way to factor the $h$ out directly.
Multiplying by the conjugate doesn't change the value, but it transforms the numerator from $\sqrt{1+h} - 1$ (a difference of roots) to $(\sqrt{1+h})^2 - 1$ (a difference of squares). The square of a root is just the thing inside, so the numerator simplifies to $(1+h) - 1 = h$. The $h$ is now naked, and cancellation works.
This same trick works for any square-root or radical function. If you see $\sqrt{u} - \sqrt{v}$, multiply by $\dfrac{\sqrt{u} + \sqrt{v}}{\sqrt{u} + \sqrt{v}}$ and the roots collapse.
For radical / square-root functions, the 5 steps now include a "step 1.5" — the conjugate multiplication:
When you see square roots in the difference quotient, the conjugate is the move. The hard part isn't the algebra — it's recognizing that you need to "unblock" the $h$ before the cancel step will work.
Use the definition (the 5 steps + conjugate) to find $f'(4)$ for $f(x) = \sqrt{x}$.
Hint: the dance is the same as Q3, but now $a = 4$ instead of $a = 1$, so $\sqrt{a} = 2$ (not $1$). The conjugate becomes $\sqrt{4+h} + 2$ instead of $\sqrt{1+h} + 1$.
Step 1: $\displaystyle f'(4) = \lim_{h \to 0} \dfrac{\sqrt{4+h} - 2}{h} \cdot \dfrac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}$
Step 2: FOIL: $(\sqrt{4+h})^2 - 2^2 = (4+h) - 4 = h$
Step 3: $f'(4) = \lim_{h \to 0} \dfrac{h}{h(\sqrt{4+h} + 2)}$
Step 4: Cancel: $\lim_{h \to 0} \dfrac{1}{\sqrt{4+h} + 2}$
Step 5: $f'(4) = \dfrac{1}{\sqrt{4} + 2} = \dfrac{1}{2 + 2} = \dfrac{1}{4} = 0.25$
Sanity check: $\sqrt{x}$ is concave down and its slope shrinks as $x$ grows. At $x = 1$ the slope is $1/2 = 0.5$; at $x = 4$ it should be smaller — $0.25$ checks out (and is exactly half). ✓
Three directions once you finish Q3: