70 lines
1022 B
Markdown
70 lines
1022 B
Markdown
$$f' = \dfrac{e^{-5x} * cos(3x)-e^{-5x}*[cos(3x)]'}{cos^2(3x)}$$
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$${ df(x)} over { dx } = { { e^{(-5x)'} } * cos( 3x) - e^{-5x} * [ cos(3x) ]' } over { cos^2(3x) } = { e^{-5x}*(-5 * cos(3x) 3sen(-3x))} over { cos^2(3x) } }$$
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## 2.
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$$
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\large
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y=x^3 \cdot(\sqrt{x} + x^2)
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$$
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$$
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y=x^3 \cdot(x^{\frac{1}{2}} + x^2)
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$$
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$$
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y=x^{\frac{3}{2}} + x^5
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$$
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$$
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y=x^{\frac{7}{2}} + x^5
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$$
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$$
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y'=\frac{7}{2} \cdot x^{\frac{5}{2}} + 5\cdot x^4
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$$
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$$
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y'=\small \dfrac{7}{2}. x^2\sqrt{x}+5x^4
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$$
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# 3.
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$f(x) = \sqrt{2(x^2 - 8)}+x$ se $x<-4$
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$f(x) = 2^{x+4}$ se $x\ge-4$
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# 4.
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**Coeficiente angular:**
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$3y = 2 - x$
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$y = \frac{2}{3} - \frac{x}{3}$
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$r_y = b-\frac{x}{3}$
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Para tingir (com tinta :) a derivada deve ser $-\frac{1}{3}$
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$f'(x) = 8x-3$
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$f'(x) = 8x-3$
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$8x-3 = -\frac{1}{3}$
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$8x = 3-\frac{1}{3}$
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$8x = \frac{8}{3}$
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$x = \frac{1}{3}$
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Encontrar o ponto de tangência
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$f(\frac{1}{3}) = 4\cdot(\frac{1}{3})^2 - 1$
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$f(\frac{1}{3}) = \frac{4}{9} - 1$
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$f(\frac{1}{3}) = -\frac{5}{9}$
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Equação da reta:
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$$y = mx+ b$$
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$
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