Question

Calcule a derivada de y=(2x^2+x+1)^3/2

Answer 1

Regra da cadeia:

$\boxed{\boxed{\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)}}$
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Sendo $f(x)=x^{3/2}$ e $g(x)=2x^{2}+x+1$, temos que

$y=(2x^{2}+x+1)^{3/2}=[g(x)]^{3/2}=f(g(x))$

Logo, derivaremos $y$ pela regra da cadeia:

$y'=f'(g(x))\cdot g'(x)$

Porém

$f'(x)=\dfrac{d}{dx}x^{3/2}=\dfrac{3}{2}\,x^{(3/2)-1}=\dfrac{3}{2}\,x^{1/2}$

e

$g'(x)=\dfrac{d}{dx}(2x^{2}+x+1)=2\cdot2x+1+0=4x+1$

logo

$y'=f'(g(x))\cdot g'(x)=f'(2x^{2}+2x+1)\cdot(4x+1)\\\\\\\boxed{\boxed{y'=\dfrac{3}{2}\,(2x^{2}+x+1)^{1/2}\cdot(4x+1)}}$

Answer 2

Resolução:
f(x) = (2x^2+x+1)^3/2
f`(x)=3(2.2x+1+0).(2x^2+x+1)^2/2 = 3(2x^2+x+1)^2(4x+1)/2

bons estudos: