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Worked Examples · Example 7

Q.Find the derivative of sin⁡x\sin x at x=0x = 0.

Mizoram MbseTextbookSubjective· 2mImportance★★★★★est
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✓ Free question

The derivative of sin⁡x\sin x at x=0x = 0 is 11. This comes from the limit definition of the derivative and the fundamental limit lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1.

The derivative of a function at a point tells us the slope of the tangent line at that point — the instantaneous rate of change. For sin⁡x\sin x at x=0x = 0, we're asking: how fast is sin⁡x\sin x changing exactly when xx is zero?

If you picture the graph of sin⁡x\sin x, it passes through the origin with a slope that looks like it might be 11 (since near x=0x = 0, sin⁡x≈x\sin x \approx x). But let's prove it properly.

  1. Start with the definition. The derivative of a function f(x)f(x) at a point x=ax = a is:

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

For f(x)=sin⁡xf(x) = \sin x and a=0a = 0, this becomes:

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

  1. Now evaluate the limit. This is the classic limit that defines the derivative of sine at zero. The key fact is:

lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1

This is not obvious from plugging in h=0h = 0 (which gives 0/00/0), but it's a standard result proved using geometry or the squeeze theorem.

Tip

A quick intuition: for very small hh, sin⁡h≈h\sin h \approx h (in radians). So sin⁡hh≈1\frac{\sin h}{h} \approx 1, and the approximation gets better as hh shrinks.

  1. Therefore, the derivative is simply 11:

f′(0)=1f'(0) = 1

Watch out

A common mistake is to think the derivative of sin⁡x\sin x is cos⁡x\cos x everywhere, then plug in x=0x = 0 to get cos⁡0=1\cos 0 = 1. That's correct here, but only because we already know the general derivative formula. The limit definition is the foundation that justifies that formula.

✓Final answer

The derivative of sin⁡x\sin x at x=0x = 0 is 1\boxed{1}.

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