Answer:
Explanation :
GIVEN THE EQUATION :
![f(x)\text{ = cos }^2(x)\text{ }](https://img.qammunity.org/2023/formulas/mathematics/college/eo54jr06p9w4ryty810uuaxrqmnv8qb0nj.png)
(i) Find the derivative of cos^2 (x)
![\begin{gathered} f^(\prime)(x)=(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}](https://img.qammunity.org/2023/formulas/mathematics/college/rh5otupf6uwu9v48cw8fdjj2pu0sww6tlr.png)
(ii) Now that we have calculated the derivative of cos^2 (x) = -sin(2x)
at x = /6 :
![\begin{gathered} f((\pi)/(6))\text{ = -sin \lparen2 * }(\pi)/(6)) \\ \text{ = -sin }(2\pi)/(6) \\ \text{ = -sin }(\pi)/(3) \\ \text{ = -0.018} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9aobvybuvclzfighl316uh10h1u9moqkma.png)
This means that our point is ( /6 ;- 0.018)
(iii) Calculate the slope of the tangent line :
m = f'( /6 )
= -sin2