Respuesta :
[tex]\bf y=yx^3-3x+2\impliedby \textit{using implicit differentiation}
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\cfrac{dy}{dx}=\cfrac{dy}{dx}x^3+y3x^2-3\implies \cfrac{dy}{dx}-\cfrac{dy}{dx}x^3=3yx^2-3
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\cfrac{dy}{dx}(1-x^3)=3yx^2-3\implies \boxed{\cfrac{dy}{dx}=\cfrac{3yx^2-3}{1-x^3}}\\\\
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\cfrac{dy}{dx}=\cfrac{3yx^2-3}{1-x^3}\impliedby \textit{using the quotient rule}
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\cfrac{dy^2}{dx^2}=\cfrac{3[2xy+x^2\frac{dy}{dx}(1-x^3)]-(3x^2y-3)(-3x^2)}{(1-x^3)^2}
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[/tex]
[tex]\bf \begin{cases} 3[2xy+x^2\frac{dy}{dx}](1-x^3)\implies \left[6xy+6x^2\cdot \frac{3yx^2-3}{1-x^3}\right](1-x^3)\\ 6xy-6x^4y+18x^4y-18x^2 \\ ----------------\\ (3x^2y-3)(-3x^2)\implies -9x^4y+9x^2\\ \qquad -(9x^4y+9x^2) \end{cases} \\\\\\ \cfrac{dy^2}{dx^2}=\cfrac{6xy-6x^4y+18x^4y-18x^2+9x^4y+9x^2}{(1-x^3)^2} \\\\\\ \boxed{\cfrac{dy^2}{dx^2}=\cfrac{6xy+21x^4y-9x^2}{(1-x^3)^2}}[/tex]
so hmmm since it's implicit differentiation, you'd get a function in x,y terms so. hmm if we check the graph of that second derivative, it seems to be asymptotic at x = -1, meaning an inflection point, from the denominator equaling 0
and it has an x-intercept at the origin, x = 0, another inflection point
now, the graph of the second derivative, from x > 0, the value for "y" is positive all the way, with a horizontal asymptote at y = 0
meaning the original function f(x) is concave UP at x > 0
now, when -1 < x < 0, the value of "y" drops to the negatives, thus is concave DOWN at -1 < x < 0
and if we pass the asymptotic inflection point of x = -1, to the left
x < -1, the value of "y" is on the positives, thus concave UP on that region
[tex]\bf \begin{cases} 3[2xy+x^2\frac{dy}{dx}](1-x^3)\implies \left[6xy+6x^2\cdot \frac{3yx^2-3}{1-x^3}\right](1-x^3)\\ 6xy-6x^4y+18x^4y-18x^2 \\ ----------------\\ (3x^2y-3)(-3x^2)\implies -9x^4y+9x^2\\ \qquad -(9x^4y+9x^2) \end{cases} \\\\\\ \cfrac{dy^2}{dx^2}=\cfrac{6xy-6x^4y+18x^4y-18x^2+9x^4y+9x^2}{(1-x^3)^2} \\\\\\ \boxed{\cfrac{dy^2}{dx^2}=\cfrac{6xy+21x^4y-9x^2}{(1-x^3)^2}}[/tex]
so hmmm since it's implicit differentiation, you'd get a function in x,y terms so. hmm if we check the graph of that second derivative, it seems to be asymptotic at x = -1, meaning an inflection point, from the denominator equaling 0
and it has an x-intercept at the origin, x = 0, another inflection point
now, the graph of the second derivative, from x > 0, the value for "y" is positive all the way, with a horizontal asymptote at y = 0
meaning the original function f(x) is concave UP at x > 0
now, when -1 < x < 0, the value of "y" drops to the negatives, thus is concave DOWN at -1 < x < 0
and if we pass the asymptotic inflection point of x = -1, to the left
x < -1, the value of "y" is on the positives, thus concave UP on that region
