The Derivative of Trigonometric and Arcus-Functions

• Example 1 - Determine the first derivative of the function \(y = 5\cdot \sin x + 3 \cdot \cos x.\)
  The first component is derived with application of the fourth basic derivation rule (derivation of the constant multipled by the variable i.e. \((c\cdot x)' = c\cdot x'\) ) and the third rule of derivation of basic functions i.e. \((\sin x)' = \cos x\). With application of these two rules the derivation of the first component of the \(y\) function can be written as: $$\left(5 \cdot \sin x \right)' = 5 \cdot \cos x. $$ The derivation of second component of the function y with respect to x is solved with application of the fourth basic derivation rule and the fourth rule of derivation of basic functions i.e. \((\cos x)' = -\sin x.\) With application of these two rules the derivation of the seonc component of the \(y\) function with respect to \(x\) can be written as: $$ \left(3 \cdot \cos x\right)' = - 3 \cdot \sin x.$$ Solution - the derivation of the function \(y\) with respect to variable \(x\) can be written as: $$ y' = \left( 5\cdot \sin x + 3 \cdot \cos x\right)' = 5 \cdot \cos x - 3 \cdot \sin x. $$
• Example 2 - Determine the first derivative of the function \(y = \tan x - \mathrm{cot} x\)
  The first component of function \(y\) is the \(\tan x\). Using the fifth rule of the derivation of basic functions i.e. \(\left(\tan x\right)' = \frac{1}{\cos^2 x}\). The second element of the \(y\) function is the \(\mathrm{cot} x\). Using the sixth rule of the derivation of basic function i.e. \(\left(\mathrm{cot}x\right)' = -\frac{1}{\sin^2 x}. \) Solution - the first derivative of the function \(y\) with respect to variable \(x\) is written as: $$ y' = \left(\tan x - \mathrm{cot} x\right)' = \frac{1}{\cos^2 x} - \left(-\frac{1}{\sin^2 x}\right) = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x\cos^2 x} $$
• Example 3 - Determine the first derivative of the function \(y = \frac{\sin x + \cos x}{\sin x - \cos x}\)
  In this case the sixth rule of basic derivation rules must be applied i.e. \(\left(\frac{u}{v}\right)' = \frac{u'v - v'u}{v^2}\). With application of the these rule the following term is obtained $$\begin{eqnarray} y' &=& \left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)'\\\nonumber y' &=& \frac{(\sin x + \cos x)' (\sin x - \cos x) - (\sin x - \cos x)'(\sin x + \cos x) }{(\sin x - \cos x)^2}\end{eqnarray} $$ Now the third and fourth rule for derivation of basic functions are applied i.e \((\sin x)' = \cos x, \quad (\cos x)' = - \sin x. \) Using this rules the followin solution is obtained $$ \begin{eqnarray}y ' &=& \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2 }{(\sin x - \cos x)^2}\\\nonumber y' &=& \frac{-(\sin^2 x - 2\sin x \cos x + \cos^2 x) - (\sin^2 x + 2 \sin x \cos x + \cos^2 x)}{(\sin x - \cos x)^2}\\ \nonumber y' &=& \frac{-2(\sin^2 x+\cos^2 x)}{(\sin x - \cos x)^2}\\ \nonumber y' &=& -\frac{2}{(\sin x - \cos x)^2}\end{eqnarray}$$
• Example 4 - Determine the first derivative of the function \(y = 2\cdot t \cdot \sin t - (t^2 - 2) \cdot \cos t\) \begin{eqnarray}\require{cancel} y' &=& 2\sin t + 2t \cos t - 2t\cos t + t^2 \sin t - 2\sin t \\\nonumber y' &=& \cancel{2\sin t} + \cancel{2t\cos t} - \cancel{2t\cos t} + t^2\sin t - \cancel{2\sin t}\\\nonumber y' &=& t^2 \sin t\end{eqnarray}
• Example 5Determine the first derivative of the function \(y = \mathrm{arctan} x + \mathrm{arccot} x\) -
  Solution - \begin{eqnarray}\require{cancel} y' &=& \frac{1}{1+x^2}- \frac{1}{1+x^2} = 0\end{eqnarray}
• Example 6-Determine the first derivative of the function \(y = x \cdot \cot x\)
  Solution - \begin{eqnarray}\require{cancel} y' &=& \cot x -\frac{x}{\sin^2 x}\\\nonumber y' &=& \frac{\cot x\sin^2 x - x}{\sin^2 x}\\\nonumber y'&=& \frac{\frac{\cos x}{\cancel{\sin x}} \sin^\cancel{2} x -x}{\sin^2 x} \\\nonumber y'&=& \frac{\sin x \cos x - x}{\sin^2 x}\end{eqnarray}
• Example 7 - Determine the first derivative of the funciton \(y= x \cdot \mathrm{arcsin} x\)
  Solution - \begin{eqnarray}y' &=& \mathrm{arcsin} x + \frac{x}{\sqrt{1-x^2}} \\\nonumber y'&=&\frac{\mathrm{arcsin} x\sqrt{1-x^2} + x}{\sqrt{1-x^2}}\end{eqnarray}
• Example 8 - Determine the first derivative of the function \(y = \frac{(1+x^2)\cdot \mathrm{arctg}x - x }{2}\)
  Solution - \begin{eqnarray}y'&=& \frac{\left((1+x^2)\mathrm{arctg} x \right)' -1}{2} \\\nonumber y' &=& \frac{(\mathrm{arctg} x + x^2 \mathrm{arctg} x)' -1}{2}\\\nonumber y'&=& \frac{\frac{1}{1+x^2} + 2x \mathrm{arctg} x + \frac{x^2}{1+x^2} - 1}{2}\\\nonumber y'&=& \frac{1}{2} \left(\frac{1+x^2-1-x^2}{1+x^2} + 2x \mathrm{arctg} x\right) \\\nonumber y' &=& x \mathrm{arctg} x \end{eqnarray}