Derivasjon av trigonometriske funksjoner

Du trenger kjerneregelen.

$$ \begin{array}{rcl}{g}^{\text{'}}\left(x\right) & =& {\left(\cos x\right)}^{\text{'}}\\ & =& {\left(\sin \left(\frac{\pi }{2}-x\right)\right)}^{\text{'}}\\ & =& \cos \left(\frac{\pi }{2}-x\right)·\left(-1\right)\\ & =& -\sin x\end{array}$$

Her må vi bruke kjerneregelen i tredje linje.

Bruk at $$ \tan x=\frac{\sin x}{\cos x}$$ og regelen for derivasjon av en brøkfunksjon. I forenklingen av resultatet trenger du sammenhengen $$ {\cos }^{2}v+{\sin }^{2}v=1$$.

(Merk at $$ {\cos }^{2}x$$ betyr $$ {\left(\cos x\right)}^2$$.)

$$ \begin{array}{rcl}{g}^{\text{'}}\left(x\right) & =& {\left(\tan x\right)}^{\text{'}}\\ & =& {\left(\frac{\sin x}{\cos x}\right)}^{\text{'}}\\ & =& \frac{\cos x·{\left(\sin x\right)}^{\text{'}}-\sin x·{\left(\cos x\right)}^{\text{'}}}{{\left(\cos x\right)}^2}\\ & =& \frac{\cos x·\cos x-\sin x·\left(-\sin x\right)}{{\cos }^{2}x}\\ & =& \frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =& \frac{1}{{\cos }^{2}x}\end{array}$$

$$ f^{\text{'}}\left(x\right)=2\cos x$$

$$ f^{\text{'}}\left(x\right)=\frac{1}{2}\cos \left(\frac{\pi }{6}-x\right)·\left(-1\right)=-\frac{1}{2}\cos \left(\frac{\pi }{6}-x\right)$$

$$ f^{\text{'}}\left(x\right)=\frac{1}{3}\left(-\sin 3x\right)·3=-\sin 3x$$

$$ f^{\text{'}}\left(x\right)=\frac{1}{2}·2\sin x·\cos x=\sin x·\cos x$$

Vi skriver om funksjonen litt.

$$ f\left(x\right)=\frac{2}{\sin 2x}=2{\left(\sin 2x\right)}^{-1}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& 2·\left(-1\right){\left(\sin 2x\right)}^{-2}·\cos 2x·2\\ & =& -\frac{4\cos 2x}{{\sin }^{2}2x}\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& 2\left(\sin x·\left(-\sin 2x\right)·2+\cos x·\cos 2x\right)\\ & =& 2\left(\cos x·\cos 2x-2\sin x·\sin 2x\right)\end{array}$$

$$ f^{\text{'}}\left(x\right)=-2\sin x·\cos x+2\cos x$$

$$ f^{\text{'}}\left(x\right)=\frac{1}{2}·\frac{1}{{\cos }^{2}2x}·2=\frac{1}{{\cos }^{2}2x}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \frac{2}{2\sqrt{x}}·\sin x+2\sqrt{x}·\cos x\\ & =& \frac{\sin x}{\sqrt{x}}+2\sqrt{x}·\cos x\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \frac{2·3\cos x-2x·\left(-3\sin x\right)}{{\left(3\cos x\right)}^2}\\ & =& \frac{6\cos x+6x\sin x}{9{\cos }^{2}x}\\ & =& \frac{2\left(\cos x+x\sin x\right)}{3{\cos }^{2}x}\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& e^x·\cos 2x+e^x·\left(-\sin 2x\right)·2\\ & =& e^x\left(\cos 2x-2\sin 2x\right)\end{array}$$

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \frac{1}{x}·\tan \pi x+\ln x·\frac{1}{{\cos }^2\pi x}·\pi \\ & =& \frac{\tan \pi x}{x}+\frac{\pi ·\ln x}{{\cos }^2\pi x}\end{array}$$

Vi regner om fra $$ v$$ målt i grader til $$ x$$ målt i radianer. Det betyr at

$$ x=v·\frac{180}{\pi }$$

Da kan vi skrive funksjonen som

$$ f\left(x\right)=\sin \left(\frac{\pi }{180}x\right)$$

Videre får vi

$$ \begin{array}{rcl}{f}^{\text{'}}\left(x\right) & =& \cos \left(\frac{\pi }{180}x\right)·\frac{\pi }{180}\\ f^{\text{'}}\left(v\right) & =& \frac{\pi }{180}\cos v\end{array}$$