Question

x×√(x^2+y^2) + tan ^-1(y/x)=1,find (dy/dx) ​

Answer 1

Answer:

dy

=

x−y

x+y

if log(x

2

+y

2

)=2tan

−1

x

y

Given:

\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}log(x

2

+y

2

)=2tan

−1

x

y

To Prove:

\frac{d y}{d x}=\frac{x+y}{x-y}

dx

dy

=

x−y

x+y

Solution:

\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}log(x

2

+y

2

)=2tan

−1

x

y

We need to find the derivative of y with respect to x.

Differentiating with respect to x on both sides.

\log \left(x^{2}+y^{2}\right)=2 \tan ^{-1} \frac{y}{x}log(x

2

+y

2

)=2tan

−1

x

y

\frac{1}{x^{2}+y^{2}}\left(2 x+2 y\left(\frac{d y}{d x}\right)\right)=\frac{2\left(\frac{x\left(\frac{d y}{d x}\right)-y(1)}{x^{2}}\right)}{\left(1+\frac{y^{2}}{x^{2}}\right)}

x

2

+y

2

1

(2x+2y(

dx

dy

))=

(1+

x

2

y

2

)

2(

x

2

x(

dx

dy

)−y(1)

)

Where we know that \frac{dy}{dx} =y_1

dx

dy

=y

1

\frac{2\left(x+y y_{1}\right)}{x^{2}+y^{2}}=\frac{2\left(x y_{1}-y\right)}{x^{2}+y^{2}}

x

2

+y

2

2(x+yy

1

)

=

x

2

+y

2

2(xy

1

−y)

Cancelling x^2+y^2x

2

+y

2

on both sides and multiplying 2 we get

2 x+2 y y_{1}=2 x y_{1}-2 y2x+2yy

1

=2xy

1

−2y

Cancelling out 2 on both sides

\begin{gathered}\begin{array}{l}{x+y y_{1}=x y_{1}-y} \\ {x+y=x y_{1}-y y_{1}}\end{array}\end{gathered}

x+yy

1

=xy

1

−y

x+y=xy

1

−yy

1

Shifting the term which has y_1y

1

on one side

\begin{gathered}\begin{array}{l}{x+y=y_{1}(x-y)} \\ {y_{1}=\frac{x+y}{x-y}}\end{array}\end{gathered}

x+y=y

1

(x−y)

y

1

=

x−y

x+y

Substituting \frac{d y}{d x}=y_{1}

dx

dy

=y

1

\bold{\frac{d y}{d x}=\frac{x+y}{x-y}}

dx

dy

=

x−y

x+y

Hope it helps you

Thank you

Answer 2

Answer:

${\huge{\mathfrak{\blue{\fbox{\fbox{\purple{Gud~Morning}}}}}}}$

Step-by-step explanation:

$\huge\rm\orange{A}$$\huge\rm\purple{N}$$\huge\rm\red{S}$$\huge\rm\blue{W}$$\huge\rm\green{E}$$\huge\rm\pink{R}$

Given :

x×√(x^2+y^2) + tan ^-1(y/x)=1,find (dy/dx)

Solution:

Differentiating both sides of the given relation with respect to x, we get,

dx

d

[log(x

2

+y

2

)]=2

dx

d

[tan

−1

(

x

y

)]

x

2

+y

2

1

×

dx

d

(x

2

+y

2

)=2×

1+(y/x)

2

1

×

dx

d

(

x

y

)

x

2

+y

2

1

[

dx

d

(x

2

)+

dx

d

(y

2

)]=2×

x

2

+y

2

x

2

[

x

2

x

dx

dy

−y×1

]

x

2

+y

2

1

[2x+2y

dx

dy

]=

x

2

+y

2

2

[x

dx

dy

−y]

2[x+y

dx

dy

]=2[x

dx

dy

−y]

x+y

dx

dy

=x

dx

dy

−y

dx

dy

(y−x)=−(x+y)

dx

dy

=

x−y

x+y

If u still have doubt then u can understand in the Image.