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Worked Examples · Example 8

Q.Find the derivative of f(x)=3f(x) = 3 at x=0x = 0 and at x=3x = 3.

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The derivative of a constant function is always zero, regardless of the point. So f′(0)=0f'(0) = 0 and f′(3)=0f'(3) = 0.

Concept First: What does "derivative at a point" mean?

The derivative of a function at a specific point, say x=ax = a, tells us the instantaneous rate of change of the function at that point. Geometrically, it's the slope of the tangent line to the graph of ff at the point (a,f(a))(a, f(a)).

For a constant function like f(x)=3f(x) = 3, the graph is a horizontal straight line at height y=3y = 3. A horizontal line has a slope of zero everywhere — it never rises or falls. So intuitively, the derivative should be zero at every point.

But let's confirm this using the formal definition.

Step-by-step using the limit definition

The derivative of ff at a point x=ax = a is defined as:

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

provided this limit exists.

  1. For x=0x = 0: Here a=0a = 0. Since f(x)=3f(x) = 3 for any input, we have f(0+h)=3f(0+h) = 3 and f(0)=3f(0) = 3. Substitute into the definition:

f′(0)=lim⁡h→03−3h=lim⁡h→00hf'(0) = \lim_{h \to 0} \frac{3 - 3}{h} = \lim_{h \to 0} \frac{0}{h}

For any h≠0h \neq 0, 0h=0\frac{0}{h} = 0. So the limit is simply 00.

f′(0)=0f'(0) = 0

  1. For x=3x = 3:

    Now a=3a = 3. Again, f(3+h)=3f(3+h) = 3 and f(3)=3f(3) = 3.

    f′(3)=lim⁡h→03−3h=lim⁡h→00h=0f'(3) = \lim_{h \to 0} \frac{3 - 3}{h} = \lim_{h \to 0} \frac{0}{h} = 0 …

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