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Implicit differentiation exercise

Try these problems to freshen up your Calculus skills\(!\)

  • If \(\log { ({ x }^{ 2 }+{ y }^{ 2 }) = 2\tan ^{ -1 }{ \left( \frac { y }{ x } \right) } }\), then show that \(\frac { dy }{ dx } =\frac { x+y }{ x-y }\).

  • If \(x\sqrt { 1+y } +y\sqrt { 1+x } =0\), then prove that \(\frac { dy }{ dx } =\frac { -1 }{ { \left( x+1 \right) }^{ 2 } }\).

  • If \(\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\tan ^{ -1 }{ a }\), then prove that \(\frac { dy }{ dx } =\frac { y }{ x }\).

  • If \(\sin { y=x\sin { \left( a+y \right) } }\), then prove that \(\frac { dy }{ dx } =\frac { \sin ^{ 2 }{ \left( a+y \right) } }{ \sin { a } }\).

  • If \({ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } \) and \({ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } }\), then prove that \(\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y }\).

Do post solutions. Thanks\(!\)

Swapnil Das

Note by Swapnil Das
6 months ago

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Third one :

\(\cos^{-1}\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}} \right) = \tan^{-1}(a) \)
\(\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = \cos(\tan^{-1}(a)) \)
Apply componendo-dividendo,
\( \dfrac{x^{2}}{y^{2}} = -c \) where c is a constant.
This, is just the equation of straight line through the origin.
\( \therefore \dfrac{dy}{dx} =(\text{Slope}) = \dfrac{y}{x} \) Vighnesh Shenoy · 6 months ago

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@Vighnesh Shenoy Fifth one :
\( x^{2} + y^{2} = t - \dfrac{1}{t} \)
Squaring,
\( x^{4} + y^{4} + 2x^{2}y^{2} = t^{2} + \dfrac{1}{t^{2}} - 2 \)
\( \therefore x^{2}y^{2} = - 1 \)
\( y^{2} = \dfrac{-1}{x^{2}} \)
Differentiate ,

\( 2y \dfrac{dy}{dx} = \dfrac{2}{x^{3}} \)
\( \therefore \dfrac{dy}{dx} = \dfrac{1}{x^{3}y} \) Vighnesh Shenoy · 6 months ago

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@Vighnesh Shenoy First one,
Differentiating,
\( \dfrac{2}{x^{2}+y^{2}}\left(x+y\dfrac{dy}{dx}\right) = 2\dfrac{x^{2}}{x^{2}+y^{2}}\left(x\dfrac{dy}{dx} - y\right) \cdot \dfrac{1}{x^{2}} \)
\( x + y \dfrac{dy}{dx} = x\dfrac{dy}{dx} - y \)
\( x + y = (x-y)\dfrac{dy}{dx} \)
\( \dfrac{dy}{dx} = \dfrac{x+y}{x-y} \) Vighnesh Shenoy · 6 months ago

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@Vighnesh Shenoy Second one :

\( x\sqrt{1+y} + y\sqrt{1+x} = 0 \)
\( x^{2}(1+y) = y^{2}(1+x) \)
\( x^{2} - y^{2} = y^{2}x - x^{2}y \)
\( (x+y)(x-y) = xy(y-x) \)
\( x+y = -xy \)
\( x = -y(1+x) \)
\( -y = \dfrac{x}{1+x} \)
\( -y = 1 - \dfrac{1}{1+x} \)
Differentiate,
\( -\dfrac{dy}{dx} = -\dfrac{-1}{(1+x)^{2}} \)
\( \dfrac{dy}{dx} = \dfrac{-1}{(1+x)^{2}} \) Vighnesh Shenoy · 6 months ago

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@Vighnesh Shenoy Fourth one :
\( \sin(y) = x\sin(a+y) \)
Differentiating,
\( \cos(y)\dfrac{dy}{dx} = \sin(a+y) + x\cos(a+y)\dfrac{dy}{dx} \)
\( \dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y) - x\cos(a+y)} \)
\( \dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y) - x(\cos(a)\cos(y) - \sin(a)\sin(y))} = \dfrac{\sin(a+y)}{\cos(y)(1-x\cos(a)) + x\sin(a)\sin(y)} \)

Frimt the original equation,
\( \sin(y) = x\sin(y)\cos(a) + x\cos(y)\sin(a) \)
\( 1 - x\cos(a) = x\sin(a)\cot(y) \)

Substituting,
\( \dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y)x\sin(a)\cot(y) + x\sin(a)\sin(y)} \)
\( \therefore \dfrac{dy}{dx} = \dfrac{\sin(a+y)\sin(y)}{x\sin(a)\left(\cos^{2}(y)+\sin^{2}(y)\right)} \)
Substituting,
\( \sin(y) = x\sin(a+y) \)
\( \dfrac{dy}{dx} = \dfrac{x\sin(a+y)\cdot \sin(a+y)}{x\sin(a)} = \dfrac{\sin^{2}(a+y)}{\sin(a)} \) Vighnesh Shenoy · 6 months ago

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@Vighnesh Shenoy Thanks, I am indebted to you. Swapnil Das · 6 months ago

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