Derivative of Sine of Function
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Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\sin u} = \cos u \dfrac {\d u} {\d x}$
where $\sin$ is the sine function and $\cos$ is the cosine function.
Proof
\(\ds \map {\frac \d {\d x} } {\sin u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\sin u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos u \frac {\d u} {\d x}\) | Derivative of Sine Function |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.14$
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $3$