Question

Qual a derivada das funções √(9-4x) e -(8/√x)?

Answer 1

a) y=√9-4x                                          b) y=-8.x*-1/2
  
dy/dx=(1/2 .(9-4x)* 1/2-1).(-4)               dy/dx=-8(-1/2).x*-1/2-1

dy/dx=-2(9-4x)*-1/2                               dy/dx=4.x*-3/2=4/√x³

dy/dx= -   1/ (9-4x)*1/2

Answer 2

$\frac{d}{dx} ( \sqrt{9-4x}) \\ \\ u=9-4x \\ \\ \frac{d}{dx} ( \sqrt{u})= \frac{d}{dx} ( \sqrt{u} ). \frac{d}{dx} (9-4x)= \\ \\ = \frac{1}{2} u^{( \frac{1}{2} -1)} .(-4)= \\ \\ = \frac{1}{2} . u^{(- \frac{1}{2}) } .(-4)= \\ \\ =\frac{1}{ \sqrt{u} } .(-4)= \\ \\ \frac{1}{ \sqrt{9-4x} } .(-4)=- \frac{2}{ \sqrt{9-4x} } \\ \\ \\ \\ \frac{d}{dx} (- \frac{8}{ \sqrt{x} } )=-8 \frac{d}{dx} ( \frac{1}{ \sqrt{x} } )=-8 \frac{d}{dx} ( x^{-\frac{1}{2} } )=-8(- \frac{1}{2} . x^{(- \frac{1}{2}-1) } )=$

$=-8(- \frac{1}{2} x^{- \frac{3}{2} } ) =4 \frac{1}{ \sqrt{x^3} }$