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Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

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

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Input all numbers as integers not as decimals.

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If you want more help click on Show steps - you will not lose any marks.

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Note that we regard $y$ as a function of $x$.

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Hence we have (using the chain rule):

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$\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$

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And , using the product rule:

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$\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

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Now differentiate both sides of the relation with respect to $x$.

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Find the two points $(0,a),\\;\\;(0,b),\\;\\;a \\lt b$ which lie on the curve given by the relation.

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

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

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(remember that $a \\lt b$).

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Hence find the equations of the tangents at the points $(0,a)$ and $(0,b)$.

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

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Equation of tangent at $(0,a)$:

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Find the gradient of the tangent at $(0,a)$.

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Hence the equation of the tangent at $(0,a)$ is:

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$y = \\;$[[1]]

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Equation of tangent at $(0,b)$:

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Find the gradient of the tangent at $(0,b)$.

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Hence the equation of the tangent at $(0,b)$ is:

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$y = \\;$[[3]]

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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"}, "name": "Simon's copy of SFY0004 Implicit 2", "statement": "

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

Note from the chain rule that $\\frac{d}{dx}(xy) = x\\frac{dy}{dx}+y\\frac{d}{dx}(x)= x\\frac{dy}{dx}+y$

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(a)

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On 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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#### (b)

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On putting $x=0$ in the relation we get:

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

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#### (c)

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First we find the tangent at the point $(0,-\\var{c})$.

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We find using the formula we found for $\\frac{dy}{dx}$ in part (a) that the gradient at  $(0,-\\var{c})$ is:

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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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As the tangent goes through the point $(0,\\var{-c})$ i.e. at $x=0,\\;\\;y=-\\var{c}$ we see that the equation of the tangent is:

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

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Next we find that the gradient at  $(0,1)$ is:

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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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As the tangent goes through the point $(0,1)$ i.e. at $x=0,\\;\\;y=1$ we see that the equation of the tangent is:

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

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