Implicit differentiation exercise

Try these problems to freshen up your Calculus skills!!

  • If log(x2+y2)=2tan1(yx)\log { ({ x }^{ 2 }+{ y }^{ 2 }) = 2\tan ^{ -1 }{ \left( \frac { y }{ x } \right) } }, then show that dydx=x+yxy\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 cos1(x2y2x2+y2)=tan1a\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 siny=xsin(a+y)\sin { y=x\sin { \left( a+y \right) } }, then prove that dydx=sin2(a+y)sina\frac { dy }{ dx } =\frac { \sin ^{ 2 }{ \left( a+y \right) } }{ \sin { a } }.

  • If x2+y2=t1t{ 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

Note by Swapnil Das
3 years, 6 months ago

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@Vighnesh Shenoy Thanks, I am indebted to you.

Swapnil Das - 3 years, 6 months ago

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Third one :

cos1(x2y2x2+y2)=tan1(a)\cos^{-1}\left( \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}} \right) = \tan^{-1}(a)
(x2y2x2+y2)=cos(tan1(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 - 3 years, 6 months ago

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Fifth one :
x2+y2=t1t x^{2} + y^{2} = t - \dfrac{1}{t}
Squaring,
x4+y4+2x2y2=t2+1t22 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 - 3 years, 6 months ago

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First one,
Differentiating,
2x2+y2(x+ydydx)=2x2x2+y2(xdydxy)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=xdydxy x + y \dfrac{dy}{dx} = x\dfrac{dy}{dx} - y
x+y=(xy)dydx x + y = (x-y)\dfrac{dy}{dx}
dydx=x+yxy \dfrac{dy}{dx} = \dfrac{x+y}{x-y}

A Former Brilliant Member - 3 years, 6 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)
x2y2=y2xx2y x^{2} - y^{2} = y^{2}x - x^{2}y
(x+y)(xy)=xy(yx) (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=111+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 - 3 years, 6 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)(1xcos(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)
1xcos(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)(cos2(y)+sin2(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)=sin2(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 - 3 years, 6 months ago

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