giúp mik bài này ạ, mik vote 5 sao cho cho hàm số f(x) = $\frac{cosx}{1+sinx}$ a)tính f'(0) , f'( $\pi$ ), f’ ($\pi$ /2), f'( $\pi$ /4) bài 2 tìm đạ

1) ta có: f'(x)= $\dfrac{cosx}{1+sinx}$$’$

= $\dfrac{cosx’.(1+sinx)-cosx.(1+sinx)’}{(1+sinx)^2}$

=  $\dfrac{-sinx.(1+sinx)-cosx.cosx}{(1+sinx)^2}$

=  $\dfrac{-sinx-sin^2x-cos^2x}{(1+sinx)^2}$

.

=> f'(0) =   $\dfrac{-sin0-sin^20-cos^20}{(1+sin0)^2}$

=  $\dfrac{-0-0-1}{(1+0)^2}$

= $-1$

.

f'(π) = =  $\dfrac{-sinπ-sin^2π-cos^2π}{(1+sinπ)^2}$

=  $\dfrac{-0-0-1}{(1+0)^2}$

= $-1$
.

f'(π/2) =  $\dfrac{-sin(π/2)-sin^2(π/2)-cos^2(π/2)}{(1+sin(π/2))^2}$

=  $\dfrac{-1-1-0}{(1+1)^2}$

=  $\dfrac{-2}{4}$

=  $\dfrac{-1}{2}$

.

f'(π/4) = $\dfrac{-sin(π/4)-sin^2(π/4)-cos^2(π/4)}{(1+sin(π/4))^2}$

$\dfrac{-\dfrac{\sqrt[]{2}}{2}-\dfrac{1}{2}-\dfrac{1}{2}}{(1+\dfrac{\sqrt[]{2}}{2})^2}$   

= $-2+\sqrt[]{2}$

.

2) $f'(1) = $ $\lim_{n \to 1} \dfrac{\sqrt[]{2x-x^2}-\sqrt[]{2x_{o}-x_{o}^2}}{x-x_{o}}$  

=  $\lim_{n \to 1} \dfrac{(\sqrt[]{2x-x^2}-\sqrt[]{2.1-1^2})(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2})}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$

=  $\lim_{n \to 1} \dfrac{(2x-x^2)-(2.1-1^2)}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$ 

=  $\lim_{n \to 1} \dfrac{-x^2+2x-1}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$ 

=  $\lim_{n \to 1} \dfrac{-(x-1)^2}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$ 

=  $\lim_{n \to 1} \dfrac{-(x-1)}{\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}}$ 

=  $\lim_{n \to 1} \dfrac{-(1-1)}{\sqrt[]{2.1-1^2}+\sqrt[]{2.1-1^2}}$ 

= $0$