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) = $ $\lim_{n \to 1} \dfrac{\sqrt[]{2x-x^2}-\sqrt[]{2x_{o}-x_{o}^2}}{x-x_{o}}$
= $\lim_{n \to 1} \dfrac{(\sqrt[]{2x-x^2}-\sqrt[]{2.1-1^2})(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2})}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$
= $\lim_{n \to 1} \dfrac{(2x-x^2)-(2.1-1^2)}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$
= $\lim_{n \to 1} \dfrac{-x^2+2x-1}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$
= $\lim_{n \to 1} \dfrac{-(x-1)^2}{(x-1)(\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}))}$
= $\lim_{n \to 1} \dfrac{-(x-1)}{\sqrt[]{2x-x^2}+\sqrt[]{2.1-1^2}}$
= $\lim_{n \to 1} \dfrac{-(1-1)}{\sqrt[]{2.1-1^2}+\sqrt[]{2.1-1^2}}$
= $0$
