Question

Calcular as seguintes derivadas:
c) y= (x+1)^2 / x^3/2
( xis mais 1 ao quadrado sobre xis elevado na 3/2)

d) y= x(2x-1) (3x+2)

Answer 1

$\boxed{y= \frac{(x+1)^2}{x^{ \frac{3}{2} }} }$

utilizando a regra do quociente
$\boxed{\boxed{( \frac{U}{V} )' = \frac{U'*V - U*V'}{V^2} }}$

$U= (x+1)^2$

derivando usando a regra da cadeia
$u^n = n*u^{n-1}*u'$

só descer o expoente ..repetir oque ta dentro do parenteses
e multiplicar pela derivada do que ta dentro do parenteses
$U' = 2*(x+1)^{2-1} * (x^{1-1}+0)\\\\ U'= 2(x+1) = 2x+2$

derivando V
$V = x^{ \frac{3}{2} }$

$V' = \frac{3}{2}*x^{ \frac{3}{2}-1 } \\\\V' = \frac{3}{2}* x^{ \frac{1}{2} }\\\\ V' = \frac{3}{2} \sqrt{x} \\\\ V'= \frac{3 \sqrt{x} }{2}$

substituindo na regra do quociente
$\boxed{\boxed{y'= \frac{2*(x+1)* (\sqrt{x^3}) - (x+1)^2 * \frac{3 \sqrt{x} }{2} }{( \sqrt{x^3} )^2 }} }$

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$y = x(2x-1)*(3x+2)\\\\y=(2x^2-x)*(3x+2)\\\\y= (2x^2*3x) + (2x^2*2)+(-x*3x) + (-x*2)\\\\y=6x^3 +4x^2 -3x^2 -2x\\\\y=6x^3+x^2-2x\\\\\\\ y'= 6*3x^{3-1}*2x^{2-1}-2*1\\\\\\ \boxed{y'=18x^2 +2x -2}$