Question

derivada de raiz de 16+(12-x)^2

Answer 1

$y=\sqrt{16+(12-x)^2}\\ \\ u=x^2-24x+150\\ \\ y=\sqrt{u}\\ \\ u=16+(12-x)^2\\ \\ \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}\\ \\ \frac{dy}{dx}=\frac{1}{2\sqrt{x}}*(2x-24)=\frac{2x-24}{2\sqrt{x}}=\frac{x-12}{\sqrt{x}}$

Answer 2

Se for y= √[16+(12-x)²]

## Regra da cadeia

y'= (1/2) *  √[16+(12-x)²]^(1/2-1)  *  [16+(12-x)²]'

y'= (1/2) *  √[16+(12-x)²]^(-1/2)  *  [0+[(12-x)²]']

y'= (1/2) *  √[16+(12-x)²]^(-1/2)  *  [2*(12-x) * (-1)]

y'= √[16+(12-x)²]^(-1/2)  *  [ x-12 ]

y'=    [ x-12 ] / {[16+(12-x)²]^(1/4) }  é a resposta