Derive the product rule d/dx[f(x)g(x)] = f'g + fg' from first principles.

Prove Product Rule for Derivatives

Question

Prove the product rule from first principles:

\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)

Solution — Step by Step

We apply the limit definition directly to the product $h(x) = f(x) \cdot g(x)$ :

h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)\,g(x + \Delta x) - f(x)\,g(x)}{\Delta x}

Nothing clever yet — just the raw definition. The trick comes in the next step.

The numerator has two functions at different points mixed together. We can’t separate them as-is. The key move is to add and subtract $f(x + \Delta x)\,g(x)$ in the numerator:

= \lim_{\Delta x \to 0} \frac{f(x + \Delta x)\,g(x + \Delta x) - f(x + \Delta x)\,g(x) + f(x + \Delta x)\,g(x) - f(x)\,g(x)}{\Delta x}

This is the entire proof. Everything after this is just clean algebra and taking limits.

Group the first two terms and the last two terms separately:

= \lim_{\Delta x \to 0} \left[ f(x + \Delta x) \cdot \frac{g(x + \Delta x) - g(x)}{\Delta x} + g(x) \cdot \frac{f(x + \Delta x) - f(x)}{\Delta x} \right]

Now each fraction is a recognizable derivative in disguise.

Split the limit across the sum (valid because both limits exist):

$\lim_{\Delta x \to 0} \dfrac{g(x + \Delta x) - g(x)}{\Delta x} = g'(x)$
$\lim_{\Delta x \to 0} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} = f'(x)$
$\lim_{\Delta x \to 0} f(x + \Delta x) = f(x)$ (since $f$ is differentiable, hence continuous)

Substituting:

h'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)

\boxed{\frac{d}{dx}[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x)}

This is the product rule, proved from scratch.

Why This Works

The product rule looks asymmetric until you see its geometric meaning. Imagine $f(x)$ and $g(x)$ as the sides of a rectangle with area $A = f \cdot g$ . When $x$ increases by $\Delta x$ , the area changes by three pieces: $\Delta f \cdot g$ (strip along one side), $f \cdot \Delta g$ (strip along the other), and $\Delta f \cdot \Delta g$ (tiny corner piece).

As $\Delta x \to 0$ , that corner piece $\Delta f \cdot \Delta g$ becomes negligible compared to the other two — it’s second-order small. What survives is exactly $f' \cdot g + f \cdot g'$ .

The algebraic proof we did above is the rigorous way of saying the same thing. The “bridge term” $f(x + \Delta x)\,g(x)$ was our way of isolating the two strips without losing any information.

Alternative Method

Some textbooks present this using logarithmic differentiation — useful when you want to verify the rule rather than prove it.

Let $h = f \cdot g$ . Take $\ln$ of both sides:

\ln h = \ln f + \ln g

Differentiate both sides with respect to $x$ :

\frac{h'}{h} = \frac{f'}{f} + \frac{g'}{g}

Multiply through by $h = fg$ :

h' = f'g + fg'

This log-differentiation shortcut is extremely useful in JEE when you have products of 3 or more functions. For $h = f \cdot g \cdot k$ , you get $\frac{h'}{h} = \frac{f'}{f} + \frac{g'}{g} + \frac{k'}{k}$ immediately — no need to apply the product rule twice.

Common Mistake

The most common error in this proof is forgetting to use the fact that $f$ is continuous at $x$ when taking $\lim_{\Delta x \to 0} f(x + \Delta x) = f(x)$ . Students often skip this step or treat it as obvious. In a JEE Advanced proof question, examiners look for this — differentiability implies continuity, and you need continuity here to complete the argument. State it explicitly: “Since $f$ is differentiable at $x$ , it is continuous there, so $\lim_{\Delta x \to 0} f(x + \Delta x) = f(x)$ .”

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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