Derivative of Cosine Function/Proof 5

Theorem

$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$

Proof

\(\ds 1\)

\(=\)

\(\ds \cos^2 x + \sin^2 x\)

Sum of Squares of Sine and Cosine

\(\ds \leadsto \ \ \)

\(\ds \map {D_x} 1\)

\(=\)

\(\ds \map {D_x} {\cos^2 x + \sin^2 x}\)

differentiating both sides

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds 2 \map {D_x} {\cos x} \cos x + 2 \sin x \cos x\)

Product Rule for Derivatives or Chain Rule for Derivatives and Derivative of Sine Function

\(\ds \leadsto \ \ \)

\(\ds \map {D_x} {\cos x}\)

\(=\)

\(\ds -\sin x\)

provided $\cos x \ne 0$

$\blacksquare$