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calumus meretrix et gladio innocentis

Implicit Differentiation Exercises Worked


{$ \color{blue}{x^2y=10} \text{ and } \color{red}{{d \over {dx}}} $}

Contents

C25-2 {$ x^2y=10 $}

Implicitly

{$$ \begin{align} x^2y&=10 \cr { d \over {dx}} \left( x^2y \right) &= { d \over {dx}}10 \cr {d \over {dx}}x^2y + x^2{ {dy} \over {dx}} &= 0 \text{ (product rule)} \cr 2xy + x^2{{dy} \over {dx}} &= 0 \cr {{dy} \over {dx}} &= {{-2xy} \over {x^2}} \cr {{dy} \over {dx}} &= -{{2y} \over x} \cr \end{align} $$}

Explicitly

{$$ \begin{align} x^2y&=10 \cr y&= {{10} \over {x^2}} \cr y&= 10x^{-2} \cr y'&= -20x^{-3} \text{ (powers rule)} \end{align} $$}

Check There is no textbook answer, so show the two results are equivalent.

{$$ \begin{align}{{dy} \over {dx}} &= -{{2y} \over x} \cr y&= {{10} \over {x^2}} \cr {{dy} \over {dx}} &= -{{2{ {{10} \over {x^2}}}} \over x} \cr {{dy} \over {dx}} &= -20x^{-3} \end{align} $$}


{$ \color{blue}{x^2 - xy =5} \text{ and } \color{red}{{d \over {dx}}} $}

C25-3 {$ x^2 - xy =5 $}

Implicitly

{$$ \begin{align} x^2 - xy &=5 \cr {d \over {dx}} \left( x^2 - xy \right) &= {d \over {dx}}5 \cr 2x - y - x{{dy} \over {dx}} &= 0 \text{ (product rule)} \cr {{dy} \over {dx}} &= {{-2x +y} \over {-x}} \cr {{dy} \over {dx}} &= {{2x - y} \over {x}}\end{align} $$}

Explicitly

{$$ \begin{align} x^2 - xy &=5 \cr y &= \frac{5 - x^2}{-x} \cr y' &= \frac{(-x)(5-x^2)' - (5 - x^2)(-x)'}{(-x)^2} \tag{quotient rule} \cr y' &= \frac{-x(-2x) - (5 - x^2)(-1)}{x^2} \cr y' &= \frac{2x^2 + 5 -x^2}{x^2} \cr y' &= \frac{x^2+5}{x^2} \cr y' &= 1 + {5 \over {x^2}} \end{align} $$}

Check

{$$ \begin{align} {{dy} \over {dx}} &= {{2x - y} \over {x}} \cr y &= \frac{5 - x^2}{-x} \cr {{dy} \over {dx}} &= {{2x - \frac{5 - x^2}{-x} } \over {x}} \cr {{dy} \over {dx}} &= {{2x^2 - \frac{5 - x^2}{-1} } \over {x^2}} \tag{multiplying by x/x} \cr {{dy} \over {dx}} &= {{2x^2 + 5 - x^2} \over {x^2}} \cr {{dy} \over {dx}} &= {{x^2 + 5} \over {x^2}} = 1 + {5 \over {x^2}} \end{align} $$}


{$ \color{blue}{2xy +x + y = 0} \text{ and } \color{red}{{d \over {dx}}} $}

C25-4 {$ 2xy +x + y = 0 $}

Implicitly

{$$ \begin{align} 2xy +x + y &= 0 \cr (2xy +x + y)' &= (0)' \cr (2xy)' + 1 + y' &= 0 \cr 2y + 2xy' + y' &= -1 \text{ (product rule)} \cr (2x +1)y' &= -(2y + 1) \cr y' &= - \frac{2y +1}{2x +1} \end{align}$$}

Explicitly

{$$ \begin{align} 2xy +x + y &= 0 \cr (2x+1)y &= -x \cr y &= \frac{-x}{2x+1} \cr y' &= { {\left( 2x+1)(-x \right) ' - \left( -x)(2x+1 \right) ' } \over {\left( 2x+1 \right)^2} }\text{ (quotient rule)} \cr y' &= {{ ( 2x+1)(-1)- ( -x)(2)} \over {\left( 2x+1 \right)^2}} \cr y' &= {{ - 2x - 1 + 2x} \over {\left( 2x+1 \right)^2}} = - {{ 1 } \over {\left( 2x+1 \right)^2}} \end{align} $$}

Check

{$$ \begin{align} y' &= - \frac{2y +1}{2x +1} \cr y &= \frac{-x}{2x+1}\cr y' &= - \frac{2\frac{-x}{2x+1} +1}{2x +1} \cr y' &= - \frac{-2x + (2x+1)}{ \left( 2x +1 \right) ^2} \cr y' &= - \frac{1}{ \left( 2x +1 \right) ^2} \end{align}$$}


{$ \color{blue}{x^2 - 4y^2 = 36} \text{ and } \color{red}{{d \over {dx}}} $}

C25-5 {$ x^2 - 4y^2 = 36 $}

Implicit {$$ \begin{align} x^2 - 4y^2 &= 36 \cr (x^2 - 4y^2)' &= (36)' \cr 2x - 4(2y)y' &= 0 \cr y' &= {{-2x} \over {-8y}} \cr y' &= {{x} \over {4y}}\end{align}$$}

Explicit

{$$ \begin{align} x^2 - 4y^2 &= 36 \cr - 4y^2 &= 36 - x^2 \cr y^2 &= {1 \over 4}(x^2 - 36) \cr y &= \pm {1 \over 2} \sqrt{x^2 - 36} \cr y &= \pm {1 \over 2} (x^2 - 36)^{ 1 \over 2} \cr y' &= \pm {1 \over 2}(2x){1 \over 2} (x^2 - 36)^{-{ 1 \over 2}} \cr y' &= \pm x{1 \over 2} {{1} \over {\sqrt{x^2 - 36}}} \cr y' &= \pm {{x} \over {2\sqrt{x^2 - 36}}} \cr \end{align}$$}

Textbook answer omits the negative root. I do not know if I have overlooked a chance to get rid of it.

Check

Working only the positive root. Clearly the same approach would work for the negative root.

{$$ \begin{align} y' &= {{x} \over {4y}} \cr y &= {1 \over 2} \sqrt{x^2 - 36} \cr y' &= {{x} \over {4 {1 \over 2} \sqrt{x^2 - 36}} } \cr y' &= {{x} \over {2 \sqrt{x^2 - 36}} } \end{align}$$}


{$ \color{blue}{x^3 - y^3 = 1} \text{ and } \color{red}{{d\over{dx}}} $}

C25-6 {$ x^3 - y^3 = 1 $}

Implicit

{$$ \begin{align} x^3 - y^3 &= 1 \cr (x^3 - y^3)' &= (1)' \cr 3x^2 - 3y^2y' &= 0 \cr - 3y^2y' &= -3x^2 \cr y' &= {{-3x^2} \over {-3y^2}} \cr y' &= {{x^2} \over {y^2}} \end{align} $$}

Explicit {$$ \begin{align} x^3 - y^3 &= 1 \cr y^3 &= x^3-1 \cr y &= (x^3-1)^{1 \over 3} \cr y' &= {1 \over 3}(x^3-1)^{-{2 \over 3}}3x^2 \cr y' &= x^2(x^3-1)^{-{2 \over 3}} \cr y' &= {{x^2} \over {\sqrt[3](x^3-1)^2}} \cr \end{align} $$}

Check

{$$ \begin{align} y' &= {{x^2} \over {y^2}} \cr y &= (x^3-1)^{1 \over 3} \cr y' &= {{x^2} \over {\left( (x^3-1)^{1 \over 3} \right)^2}} \cr y' &= {{x^2} \over {(x^3-1)^{2 \over 3}}} \cr y' &= {{x^2} \over {\sqrt[3](x^3-1)^2}} \end{align} $$}


{$ \color{red}{x^3 + y^3 = a^3} \text{ and } \color{olive}{{d \over {dx}}} $}
a = 2
graph with slider on a

C25-7 {$ x^3 + y^3 = a^3 $}

Implicit

{$$ \begin{align} x^3 + y^3 &= a^3 \cr (x^3 + y^3)' &= (a^3)' \cr 3x^2 + 3y^2y' &= 0 \cr y' &= - {{x^2} \over {y^2}} \cr \end{align}$$}

Explicit {$$ \begin{align} x^3 + y^3 &= a^3 \cr y^3 &= a^3-x^3 \cr y &= (a^3-x^3)^{1 \over 3} \cr y' &= {1 \over 3}(a^3-x^3)^{-{{2} \over 3}}(-3x^2) \cr y' &= -x^2(a^3-x^3)^{-{{2} \over 3}} \cr y' &= -{{x^2} \over {(a^3-x^3)^{2 \over 3}}} \cr \end{align}$$}

Check {$$ \begin{align} y' &= - {{x^2} \over {y^2}} \cr y &= (a^3-x^3)^{1 \over 3} \cr y' &= - {{x^2} \over {((a^3-x^3)^{1 \over 3})^2}} \cr y' &= - {{x^2} \over {(a^3-x^3)^{2 \over 3}}} \cr \end{align}$$}


{$ \color{blue}{x^4 -4y^2 = 4} \text{ and } \color{red}{{d \over{dx}}} $}

C25-8 {$ x^4 -4y^2 = 4 $}

Implicit

{$$ \begin{align} x^4 -4y^2 &= 4 \cr (x^4 -4y^2)' &= (4)' \cr 4x^3 -8yy' &= 0 \cr -8yy' &= -4x^3 \cr y' &= {{x^3} \over {2y}} \cr \end{align}$$}

Explicit

{$$ \begin{align} x^4 -4y^2 &= 4 \cr -4y^2 &= 4-x^4 \cr y^2 &= {1 \over 4}x^4 -1 \cr y &=\pm ({1 \over 4}x^4 -1)^{1 \over 2} \cr y' &=\pm {1 \over 2}({1 \over 4}x^4 -1)^{-{1 \over 2}}(x^3) \cr y' &=\pm {x^3 \over {2 \sqrt{{1 \over 4}x^4 -1}}} \cr y' &=\pm {x^3 \over { \sqrt{x^4 -4}}}\end{align}$$}

Check

{$$ \begin{align} y' &= {{x^3} \over {2y}} \cr y &=\pm ({1 \over 4}x^4 -1)^{1 \over 2} \cr y' &= {{x^3} \over {2 \left( \pm ({1 \over 4}x^4 -1)^{1 \over 2}\right)}} \cr y' &= \pm {{x^3} \over {2 \sqrt ({1 \over 4}x^4 -1)}} \cr y' &=\pm {x^3 \over { \sqrt{x^4 -4}}} \end{align}$$}


{$ \color{blue}{x^3 + y^2 - 3x = 0} \text{ and } \color{red}{{d \over {dx}}} $}

C25-9 {$ x^3 + y^2 - 3x = 0 $}

Implicit {$$ \begin{align} x^3 + y^2 - 3x &= 0 \cr (x^3 + y^2 - 3x)' &= (0)' \cr 3x^2 + 2yy' - 3 &= 0 \cr 2yy' &= 3 - 3x^2 \cr 2yy' &= 3 - 3x^2 \cr y' &= {{3(1 - x^2)} \over {2y}} \end{align} $$}

Explicit {$$ \begin{align} x^3 + y^2 - 3x &= 0 \cr y^2 &= 3x - x^3 \cr y &= \pm \sqrt{3x - x^3} \cr y &= \pm (3x - x^3)^{1 \over 2} \cr y' &= {1 \over 2} (3x - x^3)^{-{1 \over 2}}(3 - 3x^2) \cr y' &= \pm {{3(1 - x^2) } \over {2(3x - x^3)^{1 \over 2}}} \cr y' &= \pm {{3(1 - x^2) } \over {2\sqrt(3x - x^3)}} \cr \end{align} $$}

check

{$$ \begin{align} y' &= {{3(1 - x^2)} \over {2y}} \cr y &= \pm (3x - x^3)^{1 \over 2} \cr y' &= {{3(1 - x^2)} \over {2(\pm (3x - x^3)^{1 \over 2})}} \cr y' &=\pm {{3(1 - x^2)} \over {2 \sqrt(3x - x^3)}} \cr \end{align} $$}


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  5. desmos.com
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Category: Math Calculus Implicit Differentiation


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August 05, 2017

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