Para obtener las derivada de funciones trigonométricas es necesario que primero sepas que es una derivada, conozcas las fórmulas para derivar y además las identidades trigonométricas.
Gráficamente la derivada de una función es la pendiente de la recta tangente a una curva en un punto determinado de dicha curva.
Formulas para obtener la derivada de funciones trigonométricas
\begin{align} 1. \frac{d}{ {du}} ( {sen} u ) = \cos u \cdot u’ \end{align}
\begin{align} 2. \frac{d}{ {du}} ( \cos u ) =- {sen} u \cdot u’ \end{align}
\begin{align} 3. \frac{d}{ {du}} ( \tan u ) = \sec^{2} u \cdot u’ \end{align}
\begin{align} 4. \frac{d}{ {du}} ( \cot u ) = -\csc^{2} u \cdot u’ \end{align}
\begin{align} 5. \frac{d}{ {du}} ( \sec u ) = \sec u \cdot \tan u \cdot u’\end{align}
\begin{align} 6. \frac{d}{ {du}} ( \csc u ) =- \csc u \cdot \cot u \cdot u’ \end{align}
\begin{align}7. \frac{d}{ {du}} ( {arcsen} u ) = \frac{1}{\sqrt{1-u^{2}}} u’
\end{align}
\begin{align} 8. \frac{d}{ {du}} ( \arccos u ) =- \frac{1}{\sqrt{1-u^{2}}} u’ \end{align}
\begin{align} 9. \frac{d}{ {du}} ( \arctan u ) = \frac{1}{\sqrt{1+u^{2}}} u’ \end{align}
\begin{align} 10. \frac{d}{ {du}} ( {arccot} u ) =- \frac{1}{\sqrt{1+u^{2}}} u’
\end{align}
\begin{align} 11. \frac{d}{ {du}} ( {arcsec} u ) = \frac{1}{u \sqrt{u^{2} -1}}u’ \end{align}
\begin{align} 12. \frac{d}{{du}} ( {arccsc} u ) =- \frac{1}{u \sqrt{u^{2} -1}}u’ \end{align}
\begin{align} 13. \frac{d}{{du}} ( {senhu} ) = {coshu} \cdot u’ \end{align}
\begin{align} 14. \frac{d}{{du}} ( {coshu} ) = {senhu} \cdot u’ \end{align}
\begin{align} 15. \frac{d}{{du}} ( {tanhu} ) = {sech}^{2} h \cdot u’ \end{align}
\begin{align} 16. \frac{d}{{du}} ( {cothu} ) =- {csch}^{2} h \cdot u’\end{align}
\begin{align} 17. \frac{d}{{du}} ( {sechu} ) =- {sechu} \cdot
{tanhu} \cdot u’ \end{align}
\begin{align} 18. \frac{d}{{du}} ( {cschu} ) =- {cschu} \cdot {cothu} \cdot u’ \end{align}
Identidades trigonométricas
Recíprocos
\begin{align} {sen} ( A ) = \frac{1}{\csc ( A )} \csc ( A ) =
\frac{1}{{sen} ( A )} \cos ( A ) = \frac{1}{\sec ( A )} \sec ( A ) = \frac{1}{\cos ( A )} \tan ( A ) = \frac{1}{\cot ( A )} \cot ( A ) = \frac{1}{\tan ( A )}\end{align}
Pitagóricas
\begin{align} {sen}^{2} ( A ) + \cos^{2} ( A ) = 1 \end{align}
\begin{align} 1+ \tan^{2} ( A ) = \sec^{2} ( A ) \end{align}
\begin{align} 1+ \cot^{2} ( A ) = \csc^{2} ( A ) \end{align}
Ángulos negativos
\begin{align}{sen} ( -A ) =- {sen} ( A ) \end{align}
\begin{align}\cos ( -A ) = \cos ( A ) \end{align}
\begin{align}\tan ( -A ) =- \tan ( A )\end{align}
Suma de Ángulos
\begin{align}{sen} ( A+B ) = {senAcosB} + {cosAsenB} \end{align}
\begin{align} \cos ( A+B ) = {cosAcosB} – {senAsenB} \end{align}
\begin{align} \tan ( A+B ) = \frac{\tan ( A ) + \tan ( B )}{1- \tan ( A ) \tan ( B )} \end{align}
Diferencia de Ángulos
\begin{align}{sen} ( A-B ) = {senAcosB} – {cosAsenB} \end{align}
\begin{align} \cos ( A-B ) = {cosAcosB} + {senAsenB} \end{align}
\begin{align} \tan ( A-B ) = \frac{\tan ( A ) – \tan ( B )}{1+ \tan ( A ) \tan ( B )} \end{align}
Ángulos dobles
\begin{align}{sen} ( 2A ) =2 {sen} ( A ) \cos ( A ) \end{align}
\begin{align}\cos ( 2A ) = \cos^{2} ( A ) – {sen}^{2} ( A ) =1-2 {sen}^{2} ( A) =2 \cos^{2} ( A ) -1 \end{align}
\begin{align} \tan ( 2A ) = \frac{2 \tan ( A )}{1-tan^{2} ( A )} \end{align}
Productos
\begin{align} 2 {sen} ( A ) \cos ( B ) = {sen} ( A+B ) + {sen} ( A-B ) \end{align}
\begin{align}2 \cos ( A ) {sen} ( B ) = {sen} ( A+B ) – {sen} ( A-B ) \end{align}
\begin{align}2 \cos ( A ) \cos ( B ) = \cos ( A+B ) + \cos ( A-B ) \end{align}
\begin{align} 2 {sen} ( A ) {sen} ( B ) = \cos ( A+B ) – \cos ( A-B ) \end{align}
Despejes importantes
\begin{align}{sen}^{2} ( A ) = \frac{1}{2} [ 1- \cos ( 2A ) ] \end{align}
\begin{align}\cos^{2} ( A ) = \frac{1}{2} [ 1+ \cos ( 2A ) ] \end{align}
\begin{align}{sen}^{2} ( A ) =1- \cos^{2} ( A ) \end{align}
\begin{align}\cos^{2} ( A ) =1- {sen}^{2} ( A ) \end{align}
\begin{align} \tan^{2} ( A ) = \sec^{2} ( A ) -1 \end{align}
\begin{align} \cot^{2} ( A ) = \csc^{2} ( A ) -1 \end{align}
Conociendo ya las formulas para obtener la derivada de funciones trigonométricas y además las identidades trigonométricas ya podrás fácilmente resolver algunas derivadas.
