Question

qual a derivada?
d/dx=[(4x^(2)+7)^(2)*(2x^(3)+1)^(4)]

Answer 1

Sejam u e v funções quaisquer diferenciáveis:

$\frac{du}{dx}.\frac{dv}{dx}=v.\frac{du}{dx}+u.\frac{dv}{dx}\iff(u.v)'=u'.v+u.v'$

$y(u(x)),\:\:\:\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}\iff[f(u(x))]'=f'(u(x)).u'(x)$

$\frac{d}{dx}[(4x^2+7)^2(2x^3+1)^4]\\\\=(2x^3+1)^4\frac{d}{dx}(4x^2+7)^2+(4x^2+7)^2\frac{d}{dx}(2x^3+1)^4\\\\=(2x^3+1)^4\frac{d}{du}(u)^2.\frac{du}{dx}+(4x^2+7)^2\frac{d}{dv}(v)^4.\frac{dv}{dx}\\\\=(2x^3+1)^4.2u.\frac{du}{dx}+(4x^2+7)^2.4v^3.\frac{dv}{dx}\\\\=(2x^3+1)^4.2(4x^2+7)(8x)+(4x^2+7)^2.4(2x^3+1)^3(6x^2)\\\\=16x(2x^3+1)^4(4x^2+7)+24x^2(4x^2+7)^2(2x^3+1)^3\\\\=[8x(2x^3+1)(4x^2+7)][2(2x^3+1)^3+3x(4x^2+7)(2x^3+1)^2]$

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Na derivação tomei

$y(u(x))=u^2,\:\:\:u(x)=4x^2+7\\\\y(v(x))=v^4,\:\:\:v(x)=2x^3+1$