Question

Найти производную при следующем значении , с решением

Answer 1

Ответ:

$f'(t)=\dfrac{sinx-cosx-1}{(1+cosx)^{2}};$

$4\sqrt{2}-6;$

Объяснение:

$f(t)=\dfrac{1-sinx}{1+cosx};$

$f'(t)= \bigg (\dfrac{1-sinx}{1+cosx} \bigg )'=\dfrac{(1-sinx)' \cdot (1+cosx)-(1-sinx) \cdot (1+cosx)'}{(1+cosx)^{2}}=$

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

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

$f' \bigg (\dfrac{\pi}{4} \bigg )=\dfrac{sin\dfrac{\pi}{4}-cos\dfrac{\pi}{4}-1}{\bigg (1+cos\dfrac{\pi}{4} \bigg )^{2}}=\dfrac{\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}-1}{\bigg (1+\dfrac{\sqrt{2}}{2} \bigg )^{2}}=\dfrac{-1}{1^{2}+2 \cdot 1 \cdot \dfrac{\sqrt{2}}{2}+\bigg (\dfrac{\sqrt{2}}{2} \bigg )^{2}}=$

$=\dfrac{-1}{1+\sqrt{2}+\dfrac{(\sqrt{2})^{2}}{2^{2}}}=\dfrac{-1}{1+\sqrt{2}+\dfrac{2}{4}}=\dfrac{-1}{1,5+\sqrt{2}}=\dfrac{-1 \cdot (1,5-\sqrt{2})}{(1,5+\sqrt{2}) \cdot (1,5-\sqrt{2})}=$

$=\dfrac{\sqrt{2}-1,5}{(1,5)^{2}-(\sqrt{2})^{2}}=\dfrac{\sqrt{2}-1,5}{2,25-2}=\dfrac{\sqrt{2}-1,5}{0,25}=4 \cdot (\sqrt{2}-1,5)=4\sqrt{2}-6;$