Question

find dy/dx if xy = 3x^2-2y^2​

Answer 1

First, let us find  

d

y

d

x

.

x

3

+

y

3

=

1

by differentiating with respect to  

x

,

⇒

3

x

2

+

3

y

2

d

y

d

x

=

0

by subtracting  

3

x

2

,

⇒

3

y

2

d

y

d

x

=

−

3

x

2

by dividing by  

3

y

2

,

⇒

d

y

d

x

=

−

x

2

y

2

Now, let us find  

d

2

y

d

x

2

.

by differentiating with respect to  

x

,

⇒

d

2

y

d

x

2

=

−

2

x

⋅

y

2

−

x

2

⋅

2

y

d

y

d

x

(

y

2

)

2

=

−

2

x

(

y

2

−

x

y

d

y

d

x

)

y

4

by plugging in  

d

y

d

x

=

−

x

2

y

2

,

⇒

d

2

y

d

x

2

=

−

2

x

[

y

2

−

x

y

(

−

x

2

y

2

)

]

y

4

=

−

2

x

(

y

2

+

x

3

y

)

y

4

by multiplying the numerator and the denominator by  

y

,

⇒

d

2

y

d

x

2

=

−

2

x

(

y

3

+

x

3

)

y

5

by plugging in  

y

3

+

x

3

=

1

,

⇒

d

2

y

d

x

2

=

−

2

x

y

5

Answer 2

Question :-

$\bf \:If \: xy = {3x}^{2} - {2y}^{2} , \: find \: \dfrac{dy}{dx}$

Answer

Given :-

$\bf \:xy = {3x}^{2} - {2y}^{2}$

To Find :-

$\bf \:\dfrac{dy}{dx}$

Formula used :-

$\bf \:\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

$\bf \:\dfrac{d}{dx}uv = u \dfrac{d}{dx}v + v\dfrac{d}{dx}u$

Solution :-

$\bf \:xy = {3x}^{2} - {2y}^{2}$

Differentiate w. r. t. x, we get

$\bf\implies \:\dfrac{d}{dx}xy = \dfrac{d}{dx}({3x}^{2} - {2y}^{2})$

$\bf\implies \:x\dfrac{d}{dx}y + y\dfrac{d}{dx}x = 3\dfrac{d}{dx} {x}^{2} - 2\dfrac{d}{dx} {y}^{2}$

$\bf\implies \:x\dfrac{dy}{dx} + y = 6x - 4y\dfrac{dy}{dx}$

$\bf\implies \:(x + 4y)\dfrac{dy}{dx} = 6x - y$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{6x - y}{x + 4y}$