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Hint:

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Note that we regard $y$ as a function of $x$. Hence we have (using the Chain Rule): $\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$. And, using the Product Rule: $\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

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Now differentiate both sides of the relation with respect to $x$. Below is a worked solution to the problem, but only look at it if you are struggling.

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a) By differentiating both sides of the equation implicitly we get
\$2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\$
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\$( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\$ and hence on further rearranging:

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

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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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Given the following relation between $x$ and $y$
\$\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\$
Find $\\dfrac{dy}{dx}$:

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

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Input all numbers as integers not as decimals.

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Input all numbers as integers or as fractions, not as decimals.

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