Respuesta :
Answer:
Explanation :
GIVEN THE EQUATION :
[tex]f(x)\text{ = cos }^2(x)\text{ }[/tex](i) Find the derivative of cos^2 (x)
[tex]\begin{gathered} f^{\prime}(x)=\frac{d}{dx}(cos\text{ }^2(x)\text{ \rparen.... apply the chain rule } \\ \Rightarrow2\text{ cos \lparen x\rparen }\frac{d}{dx\text{ }}(cos\text{ \lparen x\rparen\rparen} \\ \Rightarrow2cos\text{ x \lparen-sinx\rparen ..... simplify } \\ \Rightarrow-sin(2x)\text{ } \\ \therefore f^{\prime}(x)\text{ = -sin\lparen2x\rparen } \\ \\ \end{gathered}[/tex](ii) Now that we have calculated the derivative of cos^2 (x) = -sin(2x)
at x = /6 :
[tex]\begin{gathered} f(\frac{\pi}{6})\text{ = -sin \lparen2 * }\frac{\pi}{6}) \\ \text{ = -sin }\frac{2\pi}{6} \\ \text{ = -sin }\frac{\pi}{3} \\ \text{ = -0.018} \end{gathered}[/tex]This means that our point is ( /6 ;- 0.018)
(iii) Calculate the slope of the tangent line :
m = f'( /6 )
= -sin2
