Question

Seja g(x) = pln (x^2 sen^2x) definida para 0 < x < p/2. Determine o valor da taxa de variação de g(x) em relação a x no instante x=x/4

Answer 1

    $g=p.\ln(u)\\\\u=v.h\\\\v=x^2\\\\h=i^2\\\\i=sen(x)$

Utilizando a regra da cadeia e do produto:

$\frac{dg}{dx}=\frac{dg}{du}.\frac{du}{dx}\\\\\frac{dg}{dx}=p.\frac{dg}{du}.\frac{d}{dx}(v.h)\\\\\frac{dg}{du}=p.\frac{dg}{du}.[\frac{dv}{dx}.h+v.\frac{dh}{dx}]\\\\\frac{dg}{dv}=p.\frac{dg}{du}[\frac{dv}{dx}.h+v.\frac{dh}{di}.\frac{di}{dx}]\\\\\frac{dg}{dv}=p.\frac{d}{du}(\ln(u)).[\frac{d}{dx}(x^2).h+v.\frac{d}{di}(i^2).\frac{d}{dx}(sen(x)]\\\\\frac{dg}{dv}=\frac{p}{u}.(2x.h+v.2i.cos(x))\\\\\frac{dg}{dx}=\frac{p}{x^2.sen^2(x)}.(2x.sen^2(x)+x^2.2sen(x).cos(x))$

Analogamente na notação linha:

$[g(u(x))]'=g'(u(x)).u'(x)=g'(u(x)).[v(x).h(i(x))]'\\\\=g'(u(x)).[v'(x).h(i(x))+v(x).h'(i(x)).i'(x)]$

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Com a expressão em mãos, basta calcular no ponto desejado x =x/4

$g'(\frac{x}{4})=\frac{p}{(\frac{x}{4})^2.sen^2(\frac{x}{4})}.(2\frac{x}{4}.sen^2(\frac{x}{4})+(\frac{x}{4})^2.2sen(\frac{x}{4}).cos(\frac{x}{4}))\\\\g'(\frac{x}{4})=\frac{p}{(\frac{x^2}{16}).sen^2(\frac{x}{4})}.(\frac{x}{2}.sen^2(\frac{x}{4})+(\frac{x^2}{16}).2sen(\frac{x}{4}).cos(\frac{x}{4}))$