How to find the derivative of sin x using the limit definition?

Conditions

• Angles are in radians.
• $x$ is a fixed real number.
• Nonzero $h$ approaches 0.
• The standard sine-ratio limit and cosine continuity are assumed prerequisites.

Reasoning, step by step

1. Substitute $f(x) = \sin x$ into the derivative definition: $f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$.
2. Apply the sum-to-product formula $\sin A - \sin B = 2\sin\frac{A-B}{2}\cos\frac{A+B}{2}$ with $A = x+h$ and $B = x$.
3. Simplify the numerator to $2\sin\frac{h}{2}\cos\frac{2x+h}{2}$.
4. Rewrite the limit expression by splitting the denominator $h$ into $2 \cdot \frac{h}{2}$: $\lim_{h \to 0} \frac{\sin\frac{h}{2}}{\frac{h}{2}} \cdot \cos\left(x + \frac{h}{2}\right)$.
5. Evaluate the first factor using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$, which equals 1.
6. Evaluate the second factor by direct substitution of $h=0$ due to the continuity of cosine, which yields $\cos x$.
7. Multiply the results to conclude that $(\sin x)' = \cos x$.

Example

The video shows the step-by-step derivation: $f'(x) = \lim_{h \to 0} \frac{2\sin\frac{h}{2}\cos\frac{2x+h}{2}}{h} = \lim_{h \to 0} \frac{\sin\frac{h}{2}}{\frac{h}{2}} \cdot \lim_{h \to 0} \cos\frac{2x+h}{2} = 1 \cdot \cos x$.

Common misconceptions

• Believing that a limit expression with a denominator of 0 is meaningless; it is $a 0/0$ indeterminate form that requires simplification.
• Applying the standard sine limit without ensuring the variable inside the sine function exactly matches the denominator.
• Assuming the differentiation point $x$ must approach 0; $x$ remains fixed while $h$ approaches 0.

Watch the explanation

Connected concepts

Explore next

• What is the sum-to-product formula for the sine function?
• What should be noted when using the important limit $\lim (\sin x)/x = 1$?
• Why does the displayed $0/0$ warn against evaluating by direct division?

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