Question

Derive f(x)= [(raiz quadrada)x] / [x + 1]

Answer 1

$\\ Sabemos\; que \;se: \\ \\ f(x) = \displaystyle\frac{g(x)}{h(x)} \\ \\ \therefore \\ \\ f'(x) = \displaystyle\frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{h^2(x)} \\ \\ Dessa\; maneira: \\ \\ f'(x) = \displaystyle\frac{ ( \sqrt{x} )' \cdot (x+1) - ( \sqrt{x} ) \cdot (x+1)' }{ (x+1)^2} \\ \\ \begin{enumerate} \\ \\ Seja U(x) = \sqrt{x} = x^{ \frac{1}{2} } \\ \\ U'(x) = \frac{1}{2} \cdot (x)^{ \frac{1}{2} -1} \\ \\ U'(x) = \frac{1}{2} \cdot (x)^{- \frac{1}{2}}$

$\\ U'(x) = \frac{1}{2 \sqrt{x} } \\ \\ Seja \; T(x) = (x+1) \\ \\ T'(x) = ( 1 + 0) = \\ \\ T'(x) = 1 \\ \\ Desse \; modo, \; a \; derivada \; fica:$

$\\ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x} } \right)(x+1) - ( \sqrt{x} ) . (1)} {(x+1)^2} \\ \\ f'(x) = \frac{\frac{x+1}{2 \sqrt{x} } - \sqrt{x} } {(x+1)^2} \\ \\ Tirando \; "mmc" \; entre \; o \; numerador: \\ \\ \frac{x+1}{2 \sqrt{x} } - \sqrt{x} = \frac{(x+1) -2 \sqrt{x} \cdot \sqrt{x} }{2 \sqrt{x} } \\ \\ = \frac{(x+1) -2 (\sqrt{x} )^2 }{2 \sqrt{x} } \\ \\ = \frac{(x+1) -2x}{2 \sqrt{x} } \\ \\ = \frac{1 -x}{2 \sqrt{x} } \\ \\ Portanto:$

$\\ f'(x) = \frac{\left(\frac{1 -x}{2 \sqrt{x} \right)}}{(x+1)^2} \\ \\ Aplicando \; divisao \; por \; fracao: \\ \\ f'(x) = \frac{1-x }{2(x+1)^2 \sqrt{x} }$