Implicit differentiation exercise

Try these problems to freshen up your Calculus skills!!

  • If log⁡(x2+y2)=2tan⁡−1(yx)\log { ({ x }^{ 2 }+{ y }^{ 2 }) = 2\tan ^{ -1 }{ \left( \frac { y }{ x } \right) } }, then show that dydx=x+yx−y\frac { dy }{ dx } =\frac { x+y }{ x-y }.

  • If x1+y+y1+x=0x\sqrt { 1+y } +y\sqrt { 1+x } =0, then prove that dydx=−1(x+1)2\frac { dy }{ dx } =\frac { -1 }{ { \left( x+1 \right) }^{ 2 } }.

  • If cos⁡−1(x2−y2x2+y2)=tan⁡−1a\cos ^{ -1 }{ \left( \frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\tan ^{ -1 }{ a }, then prove that dydx=yx\frac { dy }{ dx } =\frac { y }{ x }.

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

  • If x2+y2=t−1t{ x }^{ 2 }+{ y }^{ 2 }=t-\frac { 1 }{ t } and x4+y4=t2+1t2{ x }^{ 4 }+{ y }^{ 4 }={ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } }, then prove that dydx=1x3y\frac { dy }{ dx } =\frac { 1 }{ { x }^{ 3 }y }.

Do post solutions. Thanks!!

Swapnil Das

#Calculus

Note by Swapnil Das
5 years, 2 months ago

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1 vote

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Comments

Third one :

cos⁡−1(x2−y2x2+y2)=tan⁡−1(a)\cos^{-1}\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}} \right) = \tan^{-1}(a)
(x2−y2x2+y2)=cos⁡(tan⁡−1(a))\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) = \cos(\tan^{-1}(a))
Apply componendo-dividendo,
x2y2=−c \dfrac{x^{2}}{y^{2}} = -c where c is a constant.
This, is just the equation of straight line through the origin.
∴dydx=(Slope)=yx \therefore \dfrac{dy}{dx} =(\text{Slope}) = \dfrac{y}{x}

A Former Brilliant Member - 5 years, 2 months ago

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Fifth one :
x2+y2=t−1t x^{2} + y^{2} = t - \dfrac{1}{t}
Squaring,
x4+y4+2x2y2=t2+1t2−2 x^{4} + y^{4} + 2x^{2}y^{2} = t^{2} + \dfrac{1}{t^{2}} - 2
∴x2y2=−1 \therefore x^{2}y^{2} = - 1
y2=−1x2 y^{2} = \dfrac{-1}{x^{2}}
Differentiate ,

2ydydx=2x3 2y \dfrac{dy}{dx} = \dfrac{2}{x^{3}}
∴dydx=1x3y \therefore \dfrac{dy}{dx} = \dfrac{1}{x^{3}y}

A Former Brilliant Member - 5 years, 2 months ago

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First one,
Differentiating,
2x2+y2(x+ydydx)=2x2x2+y2(xdydx−y)⋅1x2 \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+ydydx=xdydx−y x + y \dfrac{dy}{dx} = x\dfrac{dy}{dx} - y
x+y=(x−y)dydx x + y = (x-y)\dfrac{dy}{dx}
dydx=x+yx−y \dfrac{dy}{dx} = \dfrac{x+y}{x-y}

A Former Brilliant Member - 5 years, 2 months ago

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@A Former Brilliant Member – Second one :

x1+y+y1+x=0 x\sqrt{1+y} + y\sqrt{1+x} = 0
x2(1+y)=y2(1+x) x^{2}(1+y) = y^{2}(1+x)
x2−y2=y2x−x2y x^{2} - y^{2} = y^{2}x - x^{2}y
(x+y)(x−y)=xy(y−x) (x+y)(x-y) = xy(y-x)
x+y=−xy x+y = -xy
x=−y(1+x) x = -y(1+x)
−y=x1+x -y = \dfrac{x}{1+x}
−y=1−11+x -y = 1 - \dfrac{1}{1+x}
Differentiate,
−dydx=−−1(1+x)2 -\dfrac{dy}{dx} = -\dfrac{-1}{(1+x)^{2}}
dydx=−1(1+x)2 \dfrac{dy}{dx} = \dfrac{-1}{(1+x)^{2}}

A Former Brilliant Member - 5 years, 2 months ago

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@A Former Brilliant Member – Fourth one :
sin⁡(y)=xsin⁡(a+y) \sin(y) = x\sin(a+y)
Differentiating,
cos⁡(y)dydx=sin⁡(a+y)+xcos⁡(a+y)dydx \cos(y)\dfrac{dy}{dx} = \sin(a+y) + x\cos(a+y)\dfrac{dy}{dx}
dydx=sin⁡(a+y)cos⁡(y)−xcos⁡(a+y) \dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y) - x\cos(a+y)}
dydx=sin⁡(a+y)cos⁡(y)−x(cos⁡(a)cos⁡(y)−sin⁡(a)sin⁡(y))=sin⁡(a+y)cos⁡(y)(1−xcos⁡(a))+xsin⁡(a)sin⁡(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)=xsin⁡(y)cos⁡(a)+xcos⁡(y)sin⁡(a) \sin(y) = x\sin(y)\cos(a) + x\cos(y)\sin(a)
1−xcos⁡(a)=xsin⁡(a)cot⁡(y) 1 - x\cos(a) = x\sin(a)\cot(y)

Substituting,
dydx=sin⁡(a+y)cos⁡(y)xsin⁡(a)cot⁡(y)+xsin⁡(a)sin⁡(y) \dfrac{dy}{dx} = \dfrac{\sin(a+y)}{\cos(y)x\sin(a)\cot(y) + x\sin(a)\sin(y)}
∴dydx=sin⁡(a+y)sin⁡(y)xsin⁡(a)(cos⁡2(y)+sin⁡2(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)=xsin⁡(a+y) \sin(y) = x\sin(a+y)
dydx=xsin⁡(a+y)⋅sin⁡(a+y)xsin⁡(a)=sin⁡2(a+y)sin⁡(a) \dfrac{dy}{dx} = \dfrac{x\sin(a+y)\cdot \sin(a+y)}{x\sin(a)} = \dfrac{\sin^{2}(a+y)}{\sin(a)}

A Former Brilliant Member - 5 years, 2 months ago

@Vighnesh Shenoy Thanks, I am indebted to you.

Swapnil Das - 5 years, 2 months ago
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