Derivatives of Sine and Cosine Functions (College Board AP® Calculus AB): Revision Note

Derivatives of sine and cosine functions

How do I differentiate sin x and cos x?

• It can be shown that:

  • $\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)-\sin x}{h}=\cos x$ for all real $x$

  • $\underset{x\to a}{\lim }\frac{\sin x-\sin a}{x-a}=\cos a$ for all real $a$

• Therefore, if $f\left(x\right)=\sin x$ then $f^{\text{'}}\left(x\right)=\cos x$

• It can also be shown that:

  • $\underset{h\to 0}{\lim }\frac{\cos \left(x+h\right)-\cos x}{h}=-\sin x$ for all real $x$

  • $\underset{x\to a}{\lim }\frac{\cos x-\cos a}{x-a}=-\sin a$ for all real $a$

• Therefore, if $f\left(x\right)=\cos x$ then $f^{\text{'}}\left(x\right)=-\sin x$

How do I differentiate sin kx and cos kx?

• If $f\left(x\right)=\sin kx$ then $f^{\text{'}}\left(x\right)=k\cos kx$

• If $g\left(x\right)=\cos kx$ then $g^{\text{'}}\left(x\right)=-k\sin kx$

• These occur as a result of applying the chain rule

How do I use the definition of a derivative to differentiate sin x and cos x?

• The definition of a derivative as a limit can be used to obtain the above results

  • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

• You should know the following two trigonometric addition formulae:

  • $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$

  • $\cos \left(A+B\right)=\cos A\cos B-\sin A\sin B$

• The following two trigonometric limit theorems can be derived using the squeeze theorem

  • $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

  • $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0$

• If $f\left(x\right)=\sin x$, then using the definition of a derivative,

  • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)-\sin \left(x\right)}{h}$

• Using the addition formula $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$,

  • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$

• Factoring so that the terms $\frac{\sin h}{h}$ and $\frac{\cos h-1}{h}$ are present

  • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\left(\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)\right)$

• Applying the limits $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$ and $\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0$

  • $f^{\text{'}}\left(x\right)=\sin x·\left(0\right)+\cos x·\left(1\right)$

• So it can be concluded that $f^{\text{'}}\left(x\right)=\cos x$

• The method for $\cos x$ is shown in the worked example below