Matemática, perguntado por LucasJairo, 1 ano atrás

Derivada de
y=(1-x^2)arccosx

Soluções para a tarefa

Respondido por deividsilva784
1
 \\ y = (1-x^2)arcCosx
 \\ 
 \\ y' = (1-x^2)'arcCosx + (1-x^2)*ArcCosx'
 \\ 
 \\ y' = (0-2x)arccosx+ (1-x^2)* \frac{-1}{ \sqrt{1-x^2} } 
 \\ 
 \\ y' = -2xarccosx - \frac{(1-x^2)}{ \sqrt{1-x^2} } 
 \\ 
 \\ y' = -2xArccox +  \frac{x^2-1}{ \sqrt{1-x^2} } 
 \\ 
 \\ y' =  \frac{-2x \sqrt{1-x^2}Arccox+x^2-1 }{ \sqrt{1-x^2} } 
 \\ 
 \\ y' =  \frac{x(-2 \sqrt{1-x^2} Arccosx+x)+1}{ \sqrt{1-x^2} }
Respondido por Usuário anônimo
0

\sf \displaystyle  y=\left(1-x^2\right)arccos\:x\\\\\\\frac{d}{dx}\left(\left(1-x^2\right)arccos \left(x\right)\right)\\\\\\=\frac{d}{dx}\left(1-x^2\right)arccos \left(x\right)+\frac{d}{dx}\left(arccos \left(x\right)\right)\left(1-x^2\right)\\\\\\=\left(-2x\right)arccos \left(x\right)+\left(-\frac{1}{\sqrt{1-x^2}}\right)\left(1-x^2\right)\\\\\\\to \boxed{\sf =-2x~arccos \left(x\right)-\sqrt{1-x^2}}

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