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Worked Examples · Example 5

Q.Find the derivative at x=2x = 2 of the function f(x)=3xf(x) = 3x.

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✓ Free question

The derivative of f(x)=3xf(x)=3x is constant 33 everywhere, so at x=2x=2 it is simply 33. The answer is 3\boxed{3}.

The derivative at a point tells you the instantaneous rate of change — the slope of the tangent line — at that specific xx. For a linear function like f(x)=3xf(x)=3x, the graph is a straight line with slope 33 everywhere. That means the rate of change never varies; it’s the same constant 33 at every point, including x=2x=2.

But let’s confirm this using the formal definition, because that’s what builds real understanding.

  1. Recall the definition of the derivative at a point. The derivative of ff at x=ax=a is

f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h},

provided the limit exists. This limit measures the slope of the secant line as the two points get infinitely close.

  1. Plug in a=2a=2 and f(x)=3xf(x)=3x. We have f(2)=3⋅2=6f(2) = 3 \cdot 2 = 6, and f(2+h)=3(2+h)=6+3hf(2+h) = 3(2+h) = 6 + 3h. So the difference quotient becomes

f(2+h)−f(2)h=(6+3h)−6h=3hh.\frac{f(2+h) - f(2)}{h} = \frac{(6+3h) - 6}{h} = \frac{3h}{h}.

  1. Simplify and take the limit. For h≠0h \neq 0, 3hh=3\frac{3h}{h} = 3. The expression is constant — it doesn’t depend on hh at all. Therefore,

lim⁡h→0f(2+h)−f(2)h=lim⁡h→03=3.\lim_{h \to 0} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0} 3 = 3.

Watch out

A common mistake is to think the derivative at a point requires plugging x=2x=2 into the function and then differentiating. That’s backwards: you differentiate first (find the derivative function), then evaluate at the point. Here, since f′(x)=3f'(x)=3 for all xx, evaluating at x=2x=2 gives 33 — but the limit definition shows why it works without any shortcut.

Tip

For any linear function f(x)=mx+bf(x)=mx+b, the derivative is simply mm everywhere. You can jump straight to the answer: f′(2)=3f'(2)=3. The definition just confirms it.

✓Final answer

The derivative at x=2x=2 is 3\boxed{3}.

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