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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.
$$\therefore \dfrac{dy}{dx} =(\text{Slope}) = \dfrac{y}{x}$$ · 6 months ago

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 ,

$$2y \dfrac{dy}{dx} = \dfrac{2}{x^{3}}$$
$$\therefore \dfrac{dy}{dx} = \dfrac{1}{x^{3}y}$$ · 6 months ago

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

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,
$$-\dfrac{dy}{dx} = -\dfrac{-1}{(1+x)^{2}}$$
$$\dfrac{dy}{dx} = \dfrac{-1}{(1+x)^{2}}$$ · 6 months ago

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,
$$\sin(y) = x\sin(a+y)$$
$$\dfrac{dy}{dx} = \dfrac{x\sin(a+y)\cdot \sin(a+y)}{x\sin(a)} = \dfrac{\sin^{2}(a+y)}{\sin(a)}$$ · 6 months ago

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

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