Question

Alguém ajuda resolver, por favor

Answer 1

$f(x,y)=x^2y^3+3x^4y-6x^2+2e^{xy}+sen(x^3y^3)-lnx \\ \\ f'(x)=2xy^3+12x^3y-12x+2e^{(xy)}.y+cos(x^3y^3).3x^2y^3- \frac{1}{x} \\ \\ f'(x)=2xy^3+12x^3y-12x+2ye^{(xy)}+3x^2y^3cos(x^3y^3)- \frac{1}{x}$

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$f(x,y)=x^2y^3+3x^4y-6x^2+2e^{xy}+sen(x^3y^3)-lnx \\ \\ f'(y)=x^23y^2+3x^4.1-0+2e^{(xy)}.x+cos(x^3y^3).x^33y^2-0 \\ \\ f'(y)=3x^2y^2+3x^4+2xe^{(xy)}+3x^3y^2cos(x^3y^3)$



2ª b)
$f(x,y)= \frac{2x+y^2}{xy+3y^2} \\ \\ f(y)= \frac{(2x+y^2)'(xy+3y^2)-(2x+y^2)(xy+3y^2)'}{(xy+3y^2)^2} \\ \\ f'(y)= \frac{2y(xy+3y^2)-(2x+y^2)(2xy+6y)}{(xy+3y^2)^2} \\ \\ f'(y)= \frac{2xy^3+6y^3-4x^2y-12xy-2xy^3-6y^3}{(xy+3y^2)(xy+3y^2)} \\ \\ f'(y)= \frac{-4x^2y-12xy}{x^2y^4+6xy^4+9y^4} \\ \\ f'(y)= \frac{y(-4x^2-12x)}{y^4(x^2+6x+9)} \\ \\ f'(y)= \frac{-4x^2-12x}{y^3(x^2+6x+9)} \\ \\ f'(y)= \frac{-4x(x+3)}{y^3(x+3)(x+3)} \\ \\ f'(y)= \frac{-4x}{y^3(x+3)}$