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a)

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Hay que derivar implicitamente ambos lados de la función:

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\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]
Tomar factor común $\\displaystyle\\frac{dy}{dx}$.
\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\]

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Ahora se resuelve para $\\displaystyle\\frac{dy}{dx}$

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\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]

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b)

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Se evalúa la función cuando $x=0$ y se resuelve para $y$

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\\[\\simplify{y^2+{b}y={c}} \\Rightarrow \\simplify{y^2+{b}y-{c}=0 }\\Rightarrow (y+\\var{c})(y-1)=0\\]

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Es decir $a=-\\var{c}$ and $b=1$.

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c)

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Determinación de la tangente en  $(0,-\\var{c})$.

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En la ecaución $\\frac{dy}{dx}$ determinamos la pendiente en el punto $(0,-\\var{c})$, es decir:

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\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}+{d*c}}}{\\var{b}-\\var{2*c}}=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}\\]

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Se aplica la fórmula pendiente intersección con es punto $(0,\\var{-c})$, donde $x=0,\\;\\;y=-\\var{c}$ y la pendiente hallada:

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\\[y=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}x-\\var{c}\\]

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La pendiente en el punto  $(0,1)$ es:

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\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}-{d}}}{\\var{b}+2}=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}\\]

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Empleamos la misma fórmula, pendiente intersección, es decir: 

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\\[y=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}x+1\\]

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Calcular$\\displaystyle \\frac{dy}{dx}$ usando derivación implícita, exprese la respuesta en términos de $x$ y $y$.

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$\\displaystyle \\frac{dy}{dx}= $ [[0]]

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Ejemplo de respuesta: Digite (7−2x)/(9+2y) si la respuesta es$\\dfrac{7-2x}{9+dy}$

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Calcular los dos puntos de intersección con el eje $y$, tales que $(0,a),\\;\\;(0,b)$ y $a<b$.

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$a=\\;$[[0]]

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$b=\\;$[[1]]

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(Recuerde ingresar primero el valor menor, es decir $a<b$

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Determinar las ecuaciones de las rectas tangentes en los puntos $(0,a)$ y $(0,b)$.

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Ecuación tangente en $(0,a)$:

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El valor de la pendiente en el punto $(0,a)$ es: [[0]].

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Ecuación de la recta tangente en el punto $(0,a)$ es: $y$=[[1]]

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Ecuación tangente en $(0,b)$:

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El valor de la pendiente en el punto $(0,b)$ es: [[2]].

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Ecuación de la recta tangente en el punto $(0,b)$ es: $y$: [[3]]

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Dada la función implícita en las variables $x$ y $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]

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Implicit differentiation.

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Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.

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\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "contributors": [{"name": "Marlon Arcila", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/321/"}]}]}], "contributors": [{"name": "Marlon Arcila", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/321/"}]}