Chapter 19 Partial Differentiation-II

Review this video on homogeneous functions and Euler’s theorem before working on the questions.

19.1 Exercise 10(ii)

19.1.1 Question 1

Verify Euler’s theorem for the following functions:

  1. \(u=ax^2 + 2hxy+by^2\)

The given equation is a homogeneous function of degree \(n=2\), so we expect \(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} =nu =2u\) according to Euler’s theorem.

\[\begin{equation*} \begin{split} x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}&= x(2ax + 2hy) + y(2hx+2by)\\ &= 2ax^2 + 2hxy + 2hxy + 2by^2\\ &= 2(ax^2 + 2hxy + by^2)\\ &= 2u \end{split} \end{equation*}\]

Thus, Euler’s theorem is verified.

  1. \(u=x^3 + y^3 + z^3 - 3xyz\)

This is a homogeneous function of degree \(n=3\) with three independent variables \(f(x,y,z)\). So from Euler’s theorem, we expect

\[\begin{equation*} \begin{split} x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} &= nu = 3u \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= 3x^2 - 3yz\\ \frac{\partial u}{\partial y} &= 3y^2 - 3xz\\ \frac{\partial u}{\partial z} &= 3z^2 - 3xy \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} &= x(3x^2-3yz) + y(3y^2 - 3xz) + z(3z^2 - 3xy)\\ &= 3x^3 -3xyz + 3y^3 - 3xyz + 3z^3 -3xyz\\ &= 3(x^3 + y^3 + z^3 - 3xyz)\\ &= 3u \end{split} \end{equation*}\]

Thus, Euler’s theorem verified.

  1. \(u=(x^2 + y^2)^{1/3}\)

This is a homogeneous function of degree \(n=2/3\). So we expect \(x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} =\frac{2}{3}u\).

\[\begin{equation*} \begin{split} x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= x \frac{1}{3}(x^2 + y^2)^{-2/3}\times 2x + y \frac{1}{3}(x^2 + y^2)^{-2/3}\times 2y\\ &= \frac{2}{3}x^2(x^2+y^2)^{-2/3} + \frac{2}{3}y^2(x^2 +y^2)^{-2/3}\\ &= \frac{2}{3}(x^2 +y^2)^{-2/3}(x^2+y^2)\\ &= \frac{2}{3}(x^2 +y^2)^{1/3}\\ &= \frac{2}{3}u \end{split} \end{equation*}\]

  1. \(u=x^n \tan^{-1}\left(\frac{y}{x}\right)\) (TU 2061, 2065)

Here, \(f(x,y)=u=x^n \tan^{-1}\left(\frac{y}{x}\right)\). Then \(f(\lambda x,\lambda y)=(\lambda x)^n \tan^{-1}\left(\frac{\lambda y}{\lambda x}\right)=(\lambda)^n x^n \tan^{-1}\left(\frac{y}{x}\right)=(\lambda)^n f(x,y)\).

Thus the given function is a homogeneous function of degree \(n\). So from Euler’s theorem, we expect \(x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu = n x^n \tan^{-1}\left(\frac{y}{x}\right)\).

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= \\ & x\left(x^n \frac{1}{1+(y/x)^2}\times (-y/x^2) + \tan^{-1}\left(\frac{y}{x}\right) nx^{n-1}\right) + \\y\left(x^n \frac{1}{1+(y/x)^2}\times \frac{1}{x}\right)\\ &= -\frac{y}{1+(y/x)^2}\times x^{n-1} + \tan^{-1}\left(\frac{y}{x}\right)nx^n + \\ \frac{y}{1+(y/x)^2}\times x^{n-1}\\ &= nx^n \tan^{-1}\left(\frac{y}{x}\right)\\ &=nu \end{split} \end{equation*}\]

Hence, Euler’s theorem is verified.

  1. \(u=x f\left(\frac{y}{x}\right)\)

This is a homogeneous function of degree \(1\). So we expect

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= 1u = u =x f\left(\frac{y}{x}\right) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= x f'\left(\frac{y}{x}\right)\times \left(-\frac{y}{x^2}\right) + f\left(\frac{y}{x}\right)\\ \frac{\partial u}{\partial y} &= x \left(f'\left(\frac{y}{x}\right)\times \frac{1}{x}\right) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= x \left(x f'\left(\frac{y}{x}\right)\times \left(-\frac{y}{x^2}\right) + f\left(\frac{y}{x}\right)\right) + \\ y x \left(f'\left(\frac{y}{x}\right)\times \frac{1}{x}\right)\\ &= -y f'\left(\frac{y}{x}\right) + x f\left(\frac{y}{x}\right) + y f'\left(\frac{y}{x}\right)\\ &= xf\left(\frac{y}{x}\right)\\ &=1 \times u\\ &= u \end{split} \end{equation*}\]

  1. \(u=\frac{x^{1/4} + y^{1/4}}{x^{1/5}+ y^{1/5}}\)

The function can be written as

\[\begin{equation*} \begin{split} u &=\frac{x^{1/4} + y^{1/4}}{x^{1/5}+ y^{1/5}}\\ &= \frac{x^{1/4}(1 + (y/x)^{1/4})}{x^{1/5}(1+(y/x)^{1/5})}\\ &= x^{1/20}\frac{(1 + (y/x)^{1/4})}{(1+(y/x)^{1/5})}\\ u &= x^{1/20} \phi \left(\frac{y}{x}\right) \end{split} \end{equation*}\]

Thus this is a homogeneous function of degree \(n=1/20\). So we expect

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= \frac{1}{20}u \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= x^{1/20} \phi'(y/x)(-y/x^2) + \phi (y/x) \times \left(\frac{1}{20}\right)x^{\frac{1}{20} - 1}\\ \frac{\partial u}{\partial y} &= x^{1/20}\phi'(y/x)\times \frac{1}{x} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= x \left(x^{1/20} \phi'(y/x)(-y/x^2) + \phi (y/x) \times \\ \left(\frac{1}{20}\right)x^{\frac{1}{20} - 1}\right) + y \left(x^{1/20}\phi'(y/x)\times \frac{1}{x}\right)\\ &= \frac{1}{20} x^{1/20}\phi(y/x)\\ &= \frac{1}{20}u \end{split} \end{equation*}\]

Hence, Euler’s theorem is verified.

  1. \(u=x^n \sin \left(\frac{y}{x}\right)\)

Here, \(f(x,y)= u=x^n \sin \left(\frac{y}{x}\right)\).

\(f(\lambda x, \lambda y)=(\lambda x)^n \sin \left(\frac{\lambda y}{\lambda x}\right)= (\lambda)^n x^n \sin \left(\frac{y}{x}\right) = (\lambda)^n f(x,y)\). Thus the given function is a homogeneous one with degree \(n\).

So we need to prove

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} &= nu = n x^n \sin \left(\frac{y}{x}\right) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= x^n \cos (y/x)\times (-y/x^2) + \sin (y/x)nx^{n-1}\\ \frac{\partial u}{\partial y} &= x^n \cos (y/x)\times (1/x) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} &= x (x^n \cos (y/x)\times (-y/x^2) + \sin (y/x)nx^{n-1}) \\ + y(x^n \cos (y/x)\times (1/x))\\ &= -x^{n-1}y \cos (y/x) + nx^n \sin (y/x) + \\ x^{n-1}y \cos (y/x)\\ &= n x^n \sin \left(\frac{y}{x}\right)\\ &= nu \end{split} \end{equation*}\]

Thus, Euler’s theorem is verified.

19.1.2 Question 2

  1. If \(u=\tan^{-1} \frac{x^3 + y^3}{x-y}\), \(x \neq y\), prove that \(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin 2u\). (TU 2054)

\(u\) is not a homogeneous function of \(x\) and \(y\).

Let \(z = \tan u = \frac{x^3 + y^3}{x-y}=\frac{x^3(1+(y/x)^3)}{x(1-(y/x))}=\frac{x^2(1+(y/x)^3)}{1-(y/x)}\).

Thus \(z\) is a homogeneous function of degree \(n=2\). According to Euler’s theorem,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 2z\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial u} \times \frac{\partial u}{\partial x} = \sec^2 u \frac{\partial u}{\partial x} \\ \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial u} \times \frac{\partial u}{\partial y} = \sec^2 u \frac{\partial u}{\partial y} \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 2z\\ x\left(\sec^2 u \frac{\partial u}{\partial x}\right) + y \left(\sec^2 u \frac{\partial u}{\partial y}\right) &= 2\tan u\\ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} &= \frac{2\tan u}{\sec^2 u}\\ &= 2\frac{\sin u}{\cos u} \times \cos^2 u\\ &= 2 \sin u \cos u \\ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} &= \sin 2u \end{split} \end{equation*}\]

  1. If \(u=\sin ^{-1} \frac{x^2 + y^2}{x+y}\), prove that \(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} =\tan u\). (TU 2058, 2062, 2066)

Let \(z = \sin u = \frac{x^2 + y^2}{x + y}\). This function is a homogeneous one with degree \(1\).

Now,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 1 \times z\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial u} \times \frac{\partial u}{\partial x} = \cos u \frac{\partial u}{\partial x} \\ \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial u} \times \frac{\partial u}{\partial y} = \cos u \frac{\partial u}{\partial y} \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 1 \times z\\ x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} &= \sin u\\ x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} &= \tan u \end{split} \end{equation*}\]

  1. If \(\sin v = \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\), prove that \(x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} =0\).

Let

\[\begin{equation*} \begin{split} z = \sin v &= \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ &= \frac{\sqrt{x}(1-(y/x)^{1/2})}{\sqrt{x}(1+(y/x)^{1/2})}\\ &= x^0 \frac{(1-(y/x)^{1/2})}{(1+(y/x)^{1/2})}\\ &= x^0 f\left(\frac{y}{x}\right) \end{split} \end{equation*}\]

Thus \(z\) is a homogeneous function of degree \(0\).

According to Euler’s theorem,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 0.z\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial v} \times \frac{\partial v}{\partial x} = \cos v \frac{\partial v}{\partial x}\\ \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial v} \times \frac{\partial v}{\partial y} = \cos v \frac{\partial v}{\partial y}\\ \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &=0\\ x \cos v \frac{\partial v}{\partial x} + y \cos v \frac{\partial v}{\partial y} &= 0\\ x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} &= 0 \end{split} \end{equation*}\]

Hence proved.

  1. If \(v=\log \frac{x^2 + y^2}{x + y}\), prove that \(x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = 1\).

Let \(z= e^v = \frac{x^2 + y^2}{x + y}\). Thus \(z\) is a homogeneous function of degree \(1\). Thus according to Euler’s theorem,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 1.z \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial v} \times \frac{\partial v}{\partial x} = e^v \frac{\partial v}{\partial x}\\ \frac{\partial z}{\partial y} &= \frac{\partial z}{\partial v} \times \frac{\partial v}{\partial y} = e^v \frac{\partial v}{\partial y}\\ \end{split} \end{equation*}\]

Substituting these partial derivative values in Euler’s theorem,

\[\begin{equation*} \begin{split} x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= 1.z\\ x e^v \frac{\partial v}{\partial x} + y e^v \frac{\partial v}{\partial y} &= 1.e^v\\ x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} &= 1 \end{split} \end{equation*}\]

Hence proved.

19.1.3 Question 3

  1. Find \(\frac{dz}{dt}\) if

    1. \(z=x \log y, x=t^2, y=e^t\)

    This is a composite function. So,

    \[\begin{equation*} \begin{split} \frac{dz}{dt} &= \frac{\partial z}{\partial x} \times \frac{dx}{dt} + \frac{\partial z}{\partial y} \times \frac{dy}{dt} \end{split} \end{equation*}\]

    Now,

    \[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= \log y\\ \frac{\partial z}{\partial y} &= \frac{x}{y}\\ \frac{dx}{dt} &= 2t\\ \frac{dy}{dt} &= e^t \end{split} \end{equation*}\]

    Now substituting these values in the derivative of composite function

    \[\begin{equation*} \begin{split} \frac{dz}{dt} &= \frac{\partial z}{\partial x} \times \frac{dx}{dt} + \frac{\partial z}{\partial y} \times \frac{dy}{dt}\\ &= \log y . 2t + \frac{x}{y} . e^t\\ &= t.2t + \frac{t^2}{e^t}.e^t\\ \frac{dz}{dt} &= 3t^2 \end{split} \end{equation*}\]

    1. \(z=x^2 \sin (ax+by), x=2t + 1, y=t^2 -1\)

    The given function is a composite function.

    \[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= x^2 \cos(ax+by) \times a + \sin (ax+by) \times 2x\\ \frac{\partial z}{\partial y} &= x^2 \cos(ax+by) \times b\\ \frac{dx}{dt} &= 2\\ \frac{dy}{dt} &= 2t \end{split} \end{equation*}\]

    Now substituting these values in the derivative of composite function

    \[\begin{equation*} \begin{split} \frac{dz}{dt} &= \frac{\partial z}{\partial x} \times \frac{dx}{dt} + \frac{\partial z}{\partial y} \times \frac{dy}{dt}\\ \end{split} \end{equation*}\]

    \[\begin{equation*} \begin{split} \frac{dz}{dt} &= 2(ax^2\cos(ax+by) + \\ 2x \sin (ax+by)) + 2bt x^2 \cos(ax+by)\\ &= 2x^2 \cos (ax+by)(a+bt) + \\ 4x \sin (ax+by)\\ &= 2(a+bt)(2t+1)^2 \\ \qquad \cos\{a(2t+1)+b(t^2 -1)\} + \\ \qquad 4(2t+1)\sin\{a(2t+1)+b(t^2-1)\} \end{split} \end{equation*}\]

  2. Find \(\frac{dz}{dx}\) if \(z=(y+x)e^{xy}, y=\frac{1}{x^2}\)

Here \(z=(y+x)e^{xy}\)

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial x} &= (y+x)e^{xy}.y + e^{xy}\\ \frac{\partial z}{\partial y} &= (y+x)e^{xy}.x + e^{xy}\\ \frac{dy}{dx} &= -\frac{2}{x^3} \end{split} \end{equation*}\]

So the total derivative is given by

\[\begin{equation*} \begin{split} \frac{dz}{dx} &= \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \times \frac{dy}{dx}\\ &= (y+x)e^{xy}y + e^{xy} + [(y+x)e^{xy}x + e^{xy}] \times \left(-\frac{2}{x^3}\right)\\ &= e^{xy}\left(y^2 + xy + 1 - \frac{2y}{x^2} - \frac{2}{x}-\frac{2}{x^3}\right)\\ &= e^{1/x}\left(\frac{1}{x^4} + \frac{1}{x} + 1 - \frac{2}{x^4} - \frac{2}{x} -\frac{2}{x^3}\right)\\ &= e^{1/x}\left(1- \frac{1}{x}-\frac{2}{x^3}-\frac{1}{x^4}\right) \end{split} \end{equation*}\]

19.1.4 Question 4

If \(z=f(x,y)\) and if \(x=e^u + e^{-v}, y=e^{-u} - e^v\), prove that \(\frac{\partial z}{\partial u} -\frac{\partial z}{\partial v} =x \frac{\partial z}{\partial x} -y \frac{\partial z}{\partial y}\).

Here, the function \(z\) is a composite function of two variables \(u\) and \(v\). So,

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial u} &= \frac{\partial z}{\partial x} . \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} . \frac{\partial y}{\partial u}\\ &= \frac{\partial z}{\partial x} . e^u - \frac{\partial z}{\partial y} . e^{-u}\\ \frac{\partial z}{\partial v} &= \frac{\partial z}{\partial x} . \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} . \frac{\partial y}{\partial v}\\ &= -\frac{\partial z}{\partial x} . e^{-v} - \frac{\partial z}{\partial y} . e^v \end{split} \end{equation*}\]

Now,

Subtracting \(\frac{\partial z}{\partial v}\) from \(\frac{\partial z}{\partial u}\),

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial u}-\frac{\partial z}{\partial v}&= \frac{\partial z}{\partial x} . e^u - \frac{\partial z}{\partial y} . e^{-u} + \frac{\partial z}{\partial x} . e^{-v} + \frac{\partial z}{\partial y} . e^v\\ &= \frac{\partial z}{\partial x}(e^u + e^{-v}) - \frac{\partial z}{\partial y}(e^{-u}-e^v) \end{split} \end{equation*}\]

Substituting \(x=e^u + e^{-v}, y=e^{-u} - e^v\), we have

\[\begin{equation*} \begin{split} \frac{\partial z}{\partial u} -\frac{\partial z}{\partial v} &= x \frac{\partial z}{\partial x} -y \frac{\partial z}{\partial y} \end{split} \end{equation*}\]

19.1.5 Question 5

Find \(\frac{dy}{dx}\) in the following cases

  1. \(x^3 + y^3 = 3axy\)

Let \(f(x,y) = x^3 + y^3 - 3axy=0\). Then,

\[\begin{equation*} \begin{split} f_x &= 3x^2 - 3ay\\ f_y &= 3y^2 - 3ax \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{3x^2 - 3ay}{3y^2 - 3ax}\\ \frac{dy}{dx} &= \frac{x^2 - ay}{ax-y^2} \end{split} \end{equation*}\]

  1. \(x^{2/3} + y^{2/3} = a^{2/3}\)

Let \(f(x,y)=x^{2/3} + y^{2/3} - a^{2/3}=0\). Then,

\[\begin{equation*} \begin{split} f_x &= \frac{2}{3}x^{-1/3}\\ f_y &= \frac{2}{3}y^{-1/3} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{\frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3}}\\ \frac{dy}{dx} &= -\left(\frac{y}{x}\right)^{1/3} \end{split} \end{equation*}\]

  1. \(x^y = y^x\)

Let \(f(x,y) = x^y - y^x=0\). Then taking partial derivative w.r.t \(x\) taking \(y\) as constant.

\[\begin{equation*} \begin{split} f_x &= yx^{y-1}-y^x \log y \end{split} \end{equation*}\]

Also taking partial derivative w.r.t \(y\) taking \(x\) as constant.

\[\begin{equation*} \begin{split} f_y &= x^y \log x - xy^{x-1} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{yx^{y-1}-y^x \log y}{x^y \log x - xy^{x-1}}\\ &= - \frac{\frac{y}{x}y^x -y^x \log y}{y^x\log x- \frac{x}{y}y^x}\\ &= -\frac{\frac{y}{x}-\log y}{\log x - \frac{x}{y}}\\ &= \frac{\frac{y-x\log y}{x}}{\frac{x-y\log x}{y}}\\ \frac{dy}{dx} &= \frac{y(y-x\log y)}{x(x-y\log x)} \end{split} \end{equation*}\]

  1. \(x^p y^q = (x+y)^{p+q}\)

Let \(f(x,y)=x^py^q - (x+y)^{p+q}=0\). Then

\[\begin{equation*} \begin{split} f_x &= px^{p-1}y^q - (p+q)(x+y)^{p+q-1}\\ f_y &= qx^py^{q-1} - (p+q)(x+y)^{p+q-1} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{px^{p-1}y^q - (p+q)(x+y)^{p+q-1}}{qx^py^{q-1} - (p+q)(x+y)^{p+q-1}}\\ &= -\frac{\frac{p}{x}x^py^q - (p+q)\frac{x^py^q}{x+y}}{\frac{q}{y}x^py^q-(p+q)\frac{x^py^q}{x+y}}\\ &= -\frac{\frac{p}{x}-\frac{p+q}{x+y}}{\frac{q}{y}-\frac{p+q}{x+y}}\\ &=- \frac{\frac{px+py-px-qx}{x(x+y)}}{\frac{qx+qy-py-qy}{y(x+y)}}\\ &= \frac{py-qx}{x} \times \frac{y}{py-qx}\\ \frac{dy}{dx} &= \frac{y}{x} \end{split} \end{equation*}\]

  1. \((\tan x)^y + (y)^{\tan x}=0\)

Here, \(f(x,y) = (\tan x)^y + (y)^{\tan x}=0\).

Taking partial derivative w.r.t \(x\) setting \(y\) as constant,

\[\begin{equation*} \begin{split} \frac{\partial f}{\partial x} = f_x &= y (\tan x)^{y-1} \sec^2 x + \log y . y^{\tan x}.\sec^2 x \end{split} \end{equation*}\]

Also taking partial derivative w.r.t \(y\) setting \(x\) as constant,

\[\begin{equation*} \begin{split} \frac{\partial f}{\partial y} = f_y &= (\tan x)^y \log (\tan x) + \tan x .y^{\tan x -1} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{y (\tan x)^{y-1} \sec^2 x + \log y . y^{\tan x}.\sec^2 x}{(\tan x)^y \log (\tan x) + \tan x y^{\tan x -1}}\\ \frac{dy}{dx} &= -\frac{\sec^2 x(y (\tan x)^{y-1} + y^{\tan x}\log y)}{(\tan x)^y \log (\tan x) + \tan x. y^{\tan x -1}} \end{split} \end{equation*}\]

19.1.6 Question 6

If \(x \sqrt{1-y^2} + y \sqrt{1-x^2}=a\), show that

\[\begin{equation*} \begin{split} \frac{d^2y}{dx^2} &= -\frac{a}{(1-x^2)^{3/2}} \end{split} \end{equation*}\]

The given equation is

\[\begin{equation*} \begin{split} x \sqrt{1-y^2} + y \sqrt{1-x^2} &= a \end{split} \tag{19.1} \end{equation*}\]

which can be written as \(f(x,y) = x \sqrt{1-y^2} + y \sqrt{1-x^2}-a=0\). Then

\[\begin{equation*} \begin{split} f_x &= \sqrt{1-y^2} + y.\frac{1}{2\sqrt{1-x^2}}\times (-2x)\\ &= \sqrt{1-y^2}-\frac{xy}{\sqrt{1-x^2}}\\ f_y &= \frac{x}{2\sqrt{1-y^2}}\times (-2y) + \sqrt{1-x^2}\\ &= -\frac{xy}{\sqrt{1-y^2}} + \sqrt{1-x^2} \end{split} \end{equation*}\]

Then,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{\sqrt{1-y^2}-\frac{xy}{\sqrt{1-x^2}}}{-\frac{xy}{\sqrt{1-y^2}} + \sqrt{1-x^2}}\\ &= -\frac{(\sqrt{1-y^2}\sqrt{1-x^2}-xy)}{\sqrt{1-x^2}}\times \frac{\sqrt{1-y^2}}{-xy+\sqrt{1-x^2}\sqrt{1-y^2}}\\ \frac{dy}{dx} &= -\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} \frac{d^2y}{dx^2} &= \frac{d}{dx} \left(\frac{dy}{dx}\right)\\ &= \frac{d}{dx} \left(-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\right)\\ &= -\frac{\sqrt{1-x^2}\frac{1}{2\sqrt{1-y^2}}.\left(-2y\frac{dy}{dx}\right)-\sqrt{1-y^2}\times \frac{1}{2\sqrt{1-x^2}}\times (-2x)}{(1-x^2)}\\ &= -\frac{\frac{-2y\sqrt{1-x^2}}{2\sqrt{1-y^2}}\times \left(\frac{-\sqrt{1-y^2}}{\sqrt{1-x^2}}\right)+\frac{2x\sqrt{1-y^2}}{2\sqrt{1-x^2}}}{(1-x^2)}\\ &= - \frac{y + \frac{x\sqrt{1-y^2}}{\sqrt{1-x^2}}}{(1-x^2)}\\ \frac{d^2y}{dx^2} &= -\frac{y\sqrt{1-x^2}+x\sqrt{1-y^2}}{(1-x^2)^{3/2}} \end{split} \end{equation*}\]

Substituting numerator from (19.1), we get

\[\begin{equation*} \begin{split} \frac{d^2y}{dx^2} &= -\frac{a}{(1-x^2)^{3/2}} \end{split} \end{equation*}\]