Question

If f(x) = sin x - cos ^2x, find f'(π/6

Answer 1

Answer:

$\textbf{Here is your answer}$

Explanation:

●Sol : -

Here f(x) = sin x - cos 2x

Then,

diffrentiat both sides,

f'(x) = cosx + 2sin2x

so,

f'(pi/6) = cos(pi/6) + 2 sin 2(pi/6)

= root3/2 + 2 × root3

= root 3 + 4root3/2 = 5root3/2

Answer 2

Answer:

$\sqrt{3}$

Explanation:

f(x)=sinx -cos²x

f'(x) = $\frac{d}{dx}(sinx - cos^2x)$ = cosx - 2cosx ×(-sinx) = cosx+2sinxcosx

f'(π/6) = cos(π/6) + 2 cos(π/6)sin(π/6) = $\frac{\sqrt{3}}{2} +2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2}$  =$\sqrt{3}$