Implicit Differentiation
Solved Problems
Example 11.
\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\]
Solution.
The written equation is the equation of an ellipse. Differentiating both sides and taking into account that \(y\) is a function of \(x\), we have:
\[{\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^\prime } = {\left( 1 \right)^\prime },\;\; \Rightarrow \frac{{2x}}{{{a^2}}} + \frac{{2y}}{{{b^2}}}y' = 0,\;\; \Rightarrow y'\frac{y}{{{b^2}}} = - \frac{x}{{{a^2}}},\;\; \Rightarrow y' = - \frac{{{b^2}x}}{{{a^2}y}},\;\;\text{where}\;\;y \ne 0.\]
Example 12.
\[{3^x} + {3^y} = {3^{x + y}}\]
Solution.
Differentiate both sides and solve the resulting equation for \(y':\)
\[\left( {{3^x} + {3^y}} \right)^\prime = \left( {{3^{x + y}}} \right)^\prime,\;\; \Rightarrow {3^x}\ln 3 + {3^y}\ln 3 \cdot y' = {3^{x + y}}\ln 3 \cdot {\left( {x + y} \right)^\prime },\;\; \Rightarrow {3^x} + {3^y}y' = {3^{x + y}}\left( {1 + y'} \right),\;\; \Rightarrow {3^x} + {3^y}y' = {3^{x + y}} + {3^{x + y}}y',\;\; \Rightarrow {3^y}y' - {3^{x + y}}y' = {3^{x + y}} - {3^x},\;\; \Rightarrow {3^y}y'\left( {1 - {3^x}} \right) = {3^x}\left( {{3^y} - 1} \right), \;\Rightarrow y' = - \frac{{{3^x}\left( {{3^y} - 1} \right)}}{{{3^y}\left( {{3^x} - 1} \right)}}.\]
Example 13.
Find the derivative of the astroid \[{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = {a^{\frac{2}{3}}}.\]
Solution.
Differentiate the given equation with respect to \(x:\)
\[\left( {{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}}} \right)^\prime = \left( {{a^{\frac{2}{3}}}} \right)^\prime,\;\; \Rightarrow \frac{2}{3}{x^{ - \frac{1}{3}}} + \frac{2}{3}{y^{ - \frac{1}{3}}}y' = 0,\;\; \Rightarrow \frac{1}{{\sqrt[3]{x}}} + \frac{{y'}}{{\sqrt[3]{y}}} = 0,\;\; \Rightarrow \frac{{y'}}{{\sqrt[3]{y}}} = - \frac{1}{{\sqrt[3]{x}}},\;\; \Rightarrow y' = - \sqrt[3]{{\frac{y}{x}}},\;\;\text{where}\;\;x \ne 0,\;y \ne 0.\]
Example 14.
Find the derivative \(y'\left( x \right)\) of the function that describes the general equation of a second order curve:
\(A{x^2} + 2Bxy + C{y^2} + 2Dx +\) \(2Ey + F = 0.\)
Solution.
We differentiate this equation with respect to \(x\), taking into account that \(y\) is a function of \(x:\)
\[\left( {A{x^2} + 2Bxy + C{y^2} + 2Dx + 2Ey + F} \right)^\prime = 0,\;\; \Rightarrow 2Ax + 2B\left( {x'y + xy'} \right) + 2Cyy' + 2D + 2Ey' = 0,\;\; \Rightarrow 2Ax + 2By + 2Bxy' + 2Cyy' + 2D + 2Ey' = 0,\;\; \Rightarrow Bxy' + Cyy' + Ey' = - \left( {Ax + By + D} \right),\;\; \Rightarrow y' = - \frac{{Ax + By + D}}{{Bx + Cy + E}}.\]
Example 15.
Find the value of the derivative \(y'\left( x \right)\) at \(x = 0\) for the function given by \({e^y} + xy = e.\)
Solution.
Differentiating both sides, we obtain the following relationship:
\[\left( {{e^y} + xy} \right)^\prime = \left( e \right)^\prime,\;\; \Rightarrow {e^y}y' + x'y + xy' = 0,\;\; \Rightarrow {e^y}y' + y + xy' = 0.\]
From here we find the derivative:
\[y'\left( {{e^y} + x} \right) = - y,\;\; \Rightarrow y' = - \frac{y}{{{e^y} + x}}.\]
Calculate the value of \(y\) at \(x = 0:\)
\[{e^0} + 0 \cdot y = e,\;\; \Rightarrow {e^y} = e,\;\; \Rightarrow y = 1.\]
Then the derivative at the point \(\left( {0,1} \right)\) is equal to
\[y'\left( {x = 0,y = 1} \right) = - \frac{1}{{{e^1} + 0}} = - \frac{1}{e}.\]
Example 16.
\[x\sin y + y\cos x = 0\]
Solution.
As usual, we differentiate both sides with respect to \(x:\)
\[\left( {x\sin y + y\cos x} \right)^\prime = 0,\;\; \Rightarrow \left( {x\sin y} \right)^\prime + \left( {y\cos x} \right)^\prime = 0.\]
Using the product rule for the derivative, we have:
\[x'\sin y + x{\left( {\sin y} \right)^\prime } + y'\cos x + y{\left( {\cos x} \right)^\prime } = 0,\;\; \Rightarrow 1 \cdot \sin y + x \cdot \cos y \cdot y' + y' \cdot \cos x + y \cdot \left( { - \sin x} \right) = 0,\; \Rightarrow \sin y + x\cos y \cdot y' + \cos x \cdot y' - y\sin x = 0,\;\; \Rightarrow y'\left( {x\cos y + \cos x} \right) = y\sin x - \sin y,\;\; \Rightarrow y' = \frac{{y\sin x - \sin y}}{{x\cos y + \cos x}}.\]
Example 17.
\[y = \sin \left( {x - y} \right)\]
Solution.
Differentiate both sides of the equation with respect to \(x:\)
\[y' = {\left[ {\sin \left( {x - y} \right)} \right]^\prime },\;\; \Rightarrow y' = \cos \left( {x - y} \right) \cdot {\left( {x - y} \right)^\prime },\;\; \Rightarrow y' = \cos \left( {x - y} \right) \cdot \left( {1 - y'} \right),\;\; \Rightarrow y' = \cos \left( {x - y} \right) - \cos \left( {x - y} \right) \cdot y',\;\; \Rightarrow y'\left[ {1 + \cos \left( {x - y} \right)} \right] = \cos \left( {x - y} \right)\;\; \Rightarrow y' = \frac{{\cos \left( {x - y} \right)}}{{1 + \cos \left( {x - y} \right)}},\]
where the derivative \(y'\) is defined provided that
\[1 + \cos \left( {x - y} \right) \ne 0,\;\; \Rightarrow \cos \left( {x - y} \right) \ne - 1,\;\; \Rightarrow x - y \ne \pi + 2\pi n,\;\;n \in \mathbb{Z}.\]
Example 18.
\[y = \sin \left( {x + y} \right)\]
Solution.
\[y' = {\left[ {\sin \left( {x + y} \right)} \right]^\prime },\;\; \Rightarrow y' = \cos \left( {x + y} \right) \cdot {\left( {x + y} \right)^\prime },\;\; \Rightarrow y' = \cos \left( {x + y} \right) \cdot \left( {1 + y'} \right),\;\; \Rightarrow y' = \cos \left( {x + y} \right) + \cos \left( {x + y} \right) \cdot y',\;\; \Rightarrow y'\left[ {1 - \cos \left( {x + y} \right)} \right] = \cos \left( {x + y} \right)\;\; \Rightarrow y' = \frac{{\cos \left( {x + y} \right)}}{{1 - \cos \left( {x + y} \right)}}.\]
It is seen that the derivative is defined under condition
\[1 - \cos \left( {x + y} \right) \ne 0,\;\; \Rightarrow \cos \left( {x + y} \right) \ne 1,\;\; \Rightarrow x + y \ne 2\pi n,\;\;n \in \mathbb{Z}.\]
Example 19.
\[{x^2} - \sin \left( {xy} \right) = 0\]
We differentiate each term with respect to \(x:\)
\[\left( {{x^2}} \right)^\prime - \left( {\sin \left( {xy} \right)} \right)^\prime = 0^\prime,\;\; \Rightarrow 2x - \cos \left( {xy} \right) \cdot \left( {xy} \right)^\prime = 0,\;\; \Rightarrow 2x - \cos \left( {xy} \right) \cdot \left( {y + xy^\prime} \right) = 0,\;\; \Rightarrow 2x - y\cos \left( {xy} \right) - x\cos \left( {xy} \right)y^\prime = 0.\]
Solve for \(y^\prime:\)
\[x\cos \left( {xy} \right)y^\prime = 2x - y\cos \left( {xy} \right),\;\; \Rightarrow y^\prime = \frac{{2x - y\cos \left( {xy} \right)}}{{x\cos \left( {xy} \right)}},\;\; \Rightarrow y^\prime = \frac{2}{{\cos \left( {xy} \right)}} - \frac{y}{x}.\]
Example 20.
\[{2^{\frac{x}{y}}} = \frac{{{x^2}}}{{{y^2}}}\;\left( {y \ne 0} \right).\]
Solution.
Following the pattern described above, we have:
\[\left( {{2^{\frac{x}{y}}}} \right)^\prime = \left( {\frac{{{x^2}}}{{{y^2}}}} \right)^\prime,\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot {\left( {\frac{x}{y}} \right)^\prime } = \frac{{{{\left( {{x^2}} \right)}^\prime }{y^2} - {x^2}{{\left( {{y^2}} \right)}^\prime }}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot \frac{{x'y - xy'}}{{{y^2}}} = \frac{{2x \cdot {y^2} - {x^2} \cdot 2y \cdot y'}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2 \cdot \frac{{y - xy'}}{{{y^2}}} = \frac{{2xy\left( {y - xy'} \right)}}{{{y^4}}},\;\; \Rightarrow {2^{\frac{x}{y}}}\ln 2\left( {y - xy'} \right) = \frac{{2x}}{y}\left( {y - xy'} \right),\;\; \Rightarrow \left( {{2^{\frac{x}{y}}}\ln 2 - \frac{{2x}}{y}} \right) \cdot \left( {y - xy'} \right) = 0.\]
From the last equation we find the derivative:
\[y - xy' = 0,\;\; \Rightarrow y' = \frac{y}{x},\;\;\text{where}\;\;\;x \ne 0.\]
Note that this function does not exist at \(y = 0.\) At the same time, the derivative exists neither at \(y = 0\) nor at \(x = 0.\)
Example 21.
\[{x^2} + y + \ln \left( {x + y} \right) = 0,\] where \(x + y \gt 0.\)
Solution.
Compute the derivative \(y'\) following the common pattern described above. As a result, we have:
\[\left[ {{x^2} + y + \ln \left( {x + y} \right)} \right]^\prime = 0,\;\; \Rightarrow {2x + y' + \frac{1}{{x + y}} \cdot {\left( {x + y} \right)^\prime } = 0,\;\;} \Rightarrow {2x + y' + \frac{1}{{x + y}} \cdot \left( {1 + y'} \right) = 0,\;\;} \Rightarrow {2x + y' + \frac{1}{{x + y}} + \frac{{y'}}{{x + y}} = 0,\;\;} \Rightarrow {y'\left( {1 + \frac{1}{{x + y}}} \right) = - \left( {2x + \frac{1}{{x + y}}} \right),\;\;} \Rightarrow {y' \cdot \frac{{x + y + 1}}{{x + y}} = - \frac{{2{x^2} + 2xy + 1}}{{x + y}},\;\;} \Rightarrow {y' = - \frac{{2{x^2} + 2xy + 1}}{{x + y + 1}}.}\]
Example 22.
Find the derivative at the point \(x = 2\) for the function given by the equation
\({x^2} + {y^2} + 2x - xy +\) \({ 5y - 2 = 0.}\)
Solution.
Pre-calculate the value of \(y\) at \(x = 2:\)
\[{x^2} + {y^2} + 2x - xy + 5y - 2 = 0,\;\; \Rightarrow {y^2} + 3y + 6 = 0,\;\; \Rightarrow {\left( {y + 3} \right)^2} = 0,\;\; \Rightarrow y = - 3.\]
Now we differentiate the given equation with respect to \(x:\)
\[\left( {{x^2} + {y^2} + 2x - xy + 5y - 2} \right)^\prime = 0,\;\; \Rightarrow 2x + 2yy' + 2 - \left( {x'y + xy'} \right) + 5y' = 0,\;\; \Rightarrow 2x + 2yy' + 2 - y - xy' + 5y' = 0,\;\; \Rightarrow 2yy' - xy' + 5y' = y - 2x - 2,\;\; \Rightarrow y'\left( {2y - x + 5} \right) = y - 2x - 2,\;\; \Rightarrow y' = \frac{{y - 2x - 2}}{{2y - x + 5}}.\]
Substituting the coordinates of the point \(\left( {x = 2,y = - 3} \right),\) we find the value of the derivative:
\[y'\left( {2, - 3} \right) = \frac{{ - 3 - 2 \cdot 2 - 2}}{{2 \cdot \left( { - 3} \right) - 2 + 5}} = \frac{{ - 9}}{{ - 3}} = 3.\]
Example 23.
\[\frac{y}{x} = \ln \left( {xy} \right),\] where \(xy \gt 0.\)
Solution.
Using the product, quotient and chain rules, we obtain:
\[\left( {\frac{y}{x}} \right)^\prime = \left[ {\ln \left( {xy} \right)} \right]^\prime,\;\; \Rightarrow \frac{{y'x - yx'}}{{{x^2}}} = \frac{1}{{xy}} \cdot {\left( {xy} \right)^\prime },\;\; \Rightarrow \frac{{y'x - y}}{{{x^2}}} = \frac{{x'y + xy'}}{{xy}},\;\; \Rightarrow \frac{{y'x - y}}{x} = \frac{{y + xy'}}{y}.\]
Solve this equation for \(y':\)
\[y' - \frac{y}{x} = 1 + \frac{x}{y}y',\;\; \Rightarrow y' - \frac{x}{y}y' = 1 + \frac{y}{x},\;\; \Rightarrow y'\left( {1 - \frac{x}{y}} \right) = 1 + \frac{y}{x},\;\; \Rightarrow y' \cdot \frac{{y - x}}{y} = \frac{{y + x}}{x},\;\; \Rightarrow y' = \frac{{y + x}}{x} \cdot \frac{y}{{y - x}},\;\; \Rightarrow y' = \frac{{y\left( {y + x} \right)}}{{x\left( {y - x} \right)}}.\]
assuming that \(y \ne x.\)
Example 24.
Calculate the derivative at the point \(\left( {1,1} \right)\) of the function given by the equation \[x - y = \ln \left( {xy} \right),\] where \(xy \gt 0.\)
Solution.
Differentiate both sides of the equation with respect to \(x\) and solve for \(y^\prime:\)
\[\left( {x - y} \right)^\prime = \left( {\ln \left( {xy} \right)} \right)^\prime,\;\; \Rightarrow 1 - y^\prime = \frac{1}{{xy}} \cdot \left( {y + xy^\prime} \right),\;\; \Rightarrow xy - xyy^\prime = y + xy^\prime,\;\; \Rightarrow xy - y = y^\prime\left( {xy + x} \right),\;\; \Rightarrow y^\prime = \frac{{xy - y}}{{xy + x}}.\]
Substitute the coordinates \(\left( {1,1} \right):\)
\[y^\prime\left( {1,1} \right) = \frac{{1 \cdot 1 - 1}}{{1 \cdot 1 + 1}} = 0.\]
Example 25.
Find the derivative of the function defined by the equation \[x + y = \arctan \left( {xy} \right).\]
Solution.
Differentiate both sides of the equation with respect to \(x:\)
\[{\left( {x + y} \right)^\prime } = {\left[ {\arctan \left( {xy} \right)} \right]^\prime },\;\; \Rightarrow {1 + y' = \frac{1}{{1 + {{\left( {xy} \right)}^2}}} \cdot {\left( {xy} \right)^\prime }.}\]
Using the product rule for the derivative, we can write:
\[1 + y' = \frac{1}{{1 + {{\left( {xy} \right)}^2}}} \cdot \left( {x'y + xy'} \right),\;\; \Rightarrow 1 + y' = \frac{y}{{1 + {x^2}{y^2}}} + \frac{{xy'}}{{1 + {x^2}{y^2}}}.\]
Consequently,
\[{y' - \frac{{xy'}}{{1 + {x^2}{y^2}}} = \frac{y}{{1 + {x^2}{y^2}}} - 1,\;\;} \Rightarrow {y'\left( {1 - \frac{x}{{1 + {x^2}{y^2}}}} \right) = \frac{{y - 1 - {x^2}{y^2}}}{{1 + {x^2}{y^2}}},\;\;} \Rightarrow {y' \cdot \frac{{{x^2}{y^2} - x + 1}}{{1 + {x^2}{y^2}}} = - \frac{{{x^2}{y^2} - y + 1}}{{1 + {x^2}{y^2}}},\;\;} \Rightarrow {y' = - \frac{{{x^2}{y^2} - y + 1}}{{{x^2}{y^2} - x + 1}}.}\]
Note that the derivative \(y'\) in this example exists provided that
\[{x^2}{y^2} - x + 1 \ne 0,\;\; \Rightarrow {x^2}{y^2} \ne x - 1,\;\; \Rightarrow {y^2} \ne \frac{{x - 1}}{{{x^2}}},\;\; \Rightarrow y \ne \pm \sqrt {\frac{{x - 1}}{{{x^2}}}},\;\; \Rightarrow y \ne \pm \frac{{\sqrt {x - 1} }}{{\left| x \right|}},\;\; \Rightarrow y \ne \pm \frac{{\sqrt {x - 1} }}{x}\;\;\text{and}\;\;x \gt 1.\]
Example 26.
\[x - y + \arctan y = 0\]
Solution.
Differentiate both sides with respect to \(x:\)
\[x^\prime - y^\prime + \left( {\arctan y} \right)^\prime = 0^\prime,\;\; \Rightarrow 1 - y^\prime + \frac{{y'}}{{1 + {y^2}}} = 0.\]
Solve this equation for \(y^\prime:\)
\[y^\prime\left( {1 - \frac{1}{{1 + {y^2}}}} \right) = 1,\;\; \Rightarrow y^\prime \cdot \frac{{1 + {y^2} - 1}}{{1 + {y^2}}} = 1,\;\; \Rightarrow \frac{{y^\prime{y^2}}}{{1 + {y^2}}} = 1,\;\; \Rightarrow y^\prime = \frac{{1 + {y^2}}}{{{y^2}}},\;\; \Rightarrow y^\prime = \frac{1}{{{y^2}}} + 1.\]
Example 27.
\[x + y = {e^{x - y}}\]
Solution.
First we take logarithms of both sides:
\[\ln \left( {x + y} \right) = \ln {e^{x - y}},\;\; \Rightarrow \ln \left( {x + y} \right) = x - y.\]
Differentiate this equation with respect to \(x:\)
\[\left( {\ln \left( {x + y} \right)} \right)^\prime = \left( {x - y} \right)^\prime,\;\; \Rightarrow \frac{1}{{x + y}} \cdot \left( {1 + y^\prime} \right) = 1 - y^\prime,\;\; \Rightarrow 1 + y^\prime = \left( {1 - y^\prime} \right)\left( {x + y} \right),\;\; \Rightarrow 1 + y^\prime = x - xy^\prime + y - yy^\prime,\;\; \Rightarrow y^\prime + xy^\prime + yy^\prime = x + y - 1,\;\; \Rightarrow y^\prime\left( {1 + x + y} \right) = x + y - 1,\;\; \Rightarrow y^\prime = \frac{{x + y - 1}}{{x + y + 1}}.\]
Example 28.
\[{x^y} = {y^x}\]
Solution.
First, we take logarithms of both sides of the equation:
\[\ln \left( {{x^y}} \right) = \ln \left( {{y^x}} \right),\;\; \Rightarrow y\ln x = x\ln y.\]
Here it is assumed that \(x \gt 0\) and \(y \gt 0.\) Moreover, \(x \ne 1\) and \(y \ne 1\) as bases of the exponential functions.
Differentiating both sides of the equality with respect to \(x\) and taking into account that \(y\) is a function of \(x,\) we get:
\[\left( {y\ln x} \right)^\prime = \left( {x\ln y} \right)^\prime,\;\; \Rightarrow y' \cdot \ln x + y \cdot {\left( {\ln x} \right)^\prime } = x' \cdot \ln y + x \cdot {\left( {\ln y} \right)^\prime},\;\; \Rightarrow y' \cdot \ln x + y \cdot \frac{1}{x} = 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot y',\;\; \Rightarrow y'\left( {\ln x - \frac{x}{y}} \right) = \ln y - \frac{y}{x},\;\; \Rightarrow y' = \frac{{\ln y - \frac{y}{x}}}{{\ln x - \frac{x}{y}}},\;\; \Rightarrow y' = \frac{{y\left( {x\ln y - y} \right)}}{{x\left( {y\ln x - x} \right)}}.\]
Note that in addition to the restrictions on the allowed values of \(x\) indicated above, the derivative has a discontinuity under condition
\[y\ln x - x = 0\;\;\text{or}\;\;y = \frac{x}{{\ln x}}.\]
