Question

If \[u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\], then find \[{x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}z}}{{\partial {y^2}}}\].
A. \[\left( {2\cos 2u - 1} \right)\sin 2u\]
B. \[\left( {2\cos 2u + 1} \right)\sin 2u\]
C. \[\left( {\sin 2u - 1} \right)\cos 2u\]
D. \[\left( {\sin 2u + 1} \right)\cos 2u\]

Answer 1

Hint: First we will apply reverse of inverse trigonometry ton calculate \[\tan u\]. Then calculate the partial derivative of the equation with respect to \[x\] and \[y\]. Then multiply \[x\] with partial derivative \[x\] and \[y\] with the partial derivative of \[y\] and add these two equations. The above procedure will be repeated once again to find \[{x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}z}}{{\partial {y^2}}}\].

Formula used:
Inverse trigonometry: \[{\tan ^{ - 1}}y = x \Rightarrow y = \tan x\]
\[\dfrac{\partial }{{\partial x}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}\]
\[\dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}\]
\[\dfrac{\partial }{{\partial x}}\left( {uv} \right) = v\dfrac{{\partial u}}{{\partial x}} + u\dfrac{{\partial v}}{{\partial x}}\]

Complete step by step solution:
Given equation is
\[u = {\tan ^{ - 1}}\left( {\dfrac{{{x^3} + {y^3}}}{{x - y}}} \right)\]
Apply the reverse of inverse trigonometry:
\[ \Rightarrow \tan u = \dfrac{{{x^3} + {y^3}}}{{x - y}}\]
\[ \Rightarrow \left( {x - y} \right)\tan u = {x^3} + {y^3}\]
Compute the partial derivative with respect to \[x\]
\[ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {\left( {x - y} \right)\tan u} \right] = \dfrac{\partial }{{\partial x}}\left[ {{x^3} + {y^3}} \right]\]
Apply the product formula on the left side
\[ \Rightarrow \tan u\dfrac{\partial }{{\partial x}}\left( {x - y} \right) + \left( {x - y} \right)\dfrac{\partial }{{\partial x}}\tan u = \dfrac{\partial }{{\partial x}}\left[ {{x^3} + {y^3}} \right]\]
\[ \Rightarrow \tan u + \left( {x - y} \right){\sec ^2}u\dfrac{{\partial u}}{{\partial x}} = 3{x^2}\] …..(i)
Compute the partial derivative \[\left( {x - y} \right)\tan u = {x^3} + {y^3}\] with respect to \[y\]
\[ \Rightarrow \dfrac{\partial }{{\partial y}}\left[ {\left( {x - y} \right)\tan u} \right] = \dfrac{\partial }{{\partial y}}\left[ {{x^3} + {y^3}} \right]\]
Apply the product formula on the left side
\[ \Rightarrow \tan u\dfrac{\partial }{{\partial y}}\left( {x - y} \right) + \left( {x - y} \right)\dfrac{\partial }{{\partial y}}\tan u = \dfrac{\partial }{{\partial y}}\left[ {{x^3} + {y^3}} \right]\]
\[ \Rightarrow - \tan u + \left( {x - y} \right){\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = 3{y^2}\] ….(ii)
Multiply \[x\] with respect to (i) and multiply \[y\] with respect to (ii) and add them
\[x\tan u + y\left( {x - y} \right){\sec ^2}u\dfrac{{\partial u}}{{\partial x}} - y\tan u + y\left( {x - y} \right){\sec ^2}u\dfrac{{\partial u}}{{\partial y}} = 3{x^3} + 3{y^3}\]
\[ \Rightarrow \left( {x - y} \right)\tan u + \left( {x - y} \right){\sec ^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = 3\left( {{x^3} + {y^3}} \right)\]
Common \[\left( {x - y} \right)\] from the left side expression
\[ \Rightarrow \left( {x - y} \right)\left[ {\tan u + {{\sec }^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right)} \right] = 3\left( {{x^3} + {y^3}} \right)\]
Divide both sides by \[\left( {x - y} \right)\]
\[ \Rightarrow \tan u + {\sec ^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{{3\left( {{x^3} + {y^3}} \right)}}{{\left( {x - y} \right)}}\]
Substitute \[\tan u = \dfrac{{{x^3} + {y^3}}}{{x - y}}\] in right side expression
\[ \Rightarrow \tan u + {\sec ^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = 3\tan u\]
Subtract like terms
\[ \Rightarrow {\sec ^2}u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = 2\tan u\]
Divide both sides by \[{\sec ^2}u\]
\[ \Rightarrow \left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{{2\tan u}}{{{{\sec }^2}u}}\]
Rewrite \[\tan u = \dfrac{{\sin u}}{{\cos u}}\] and \[\sec u = \dfrac{1}{{\cos u}}\]
\[ \Rightarrow \left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{{2\dfrac{{\sin u}}{{\cos u}}}}{{\dfrac{1}{{{{\cos }^2}u}}}}\]
\[ \Rightarrow \left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = 2\sin u\cos u\]
Apply double angle formula
\[ \Rightarrow \left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \sin 2u\]
Compute the partial derivative of the above equation with respect to \[x\]
\[\dfrac{\partial }{{\partial x}}\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{\partial }{{\partial x}}\sin 2u\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{\partial u}}{{\partial x}}\dfrac{{\partial x}}{{\partial x}} + y\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + \dfrac{{\partial u}}{{\partial y}}\dfrac{{\partial y}}{{\partial x}} = 2\cos 2u\dfrac{{\partial u}}{{\partial x}}\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{\partial u}}{{\partial x}} + y\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + \dfrac{{\partial u}}{{\partial y}} \cdot 0 = 2\cos 2u\dfrac{{\partial u}}{{\partial x}}\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + \dfrac{{\partial u}}{{\partial x}} + y\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} = 2\cos 2u\dfrac{{\partial u}}{{\partial x}}\] ….(iii)
Compute partial derivative of the equation \[\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \sin 2u\] with respect to \[y\].
\[\dfrac{\partial }{{\partial y}}\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{\partial }{{\partial y}}\sin 2u\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + \dfrac{{\partial u}}{{\partial x}} \cdot \dfrac{{\partial x}}{{\partial y}} + y\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{\partial u}}{{\partial y}} \cdot \dfrac{{\partial y}}{{\partial y}} = 2\cos 2u\dfrac{{\partial u}}{{\partial y}}\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + \dfrac{{\partial u}}{{\partial x}} \cdot 0 + y\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{\partial u}}{{\partial y}} \cdot 1 = 2\cos 2u\dfrac{{\partial u}}{{\partial y}}\]
\[ \Rightarrow x\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + y\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \dfrac{{\partial u}}{{\partial y}} = 2\cos 2u\dfrac{{\partial u}}{{\partial y}}\] …..(iv)
Multiply \[x\] with respect to (iii) and multiply \[y\] with respect to (iv) and add them
\[{x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + x\dfrac{{\partial u}}{{\partial x}} + xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + y\dfrac{{\partial u}}{{\partial y}} = 2x\cos 2u\dfrac{{\partial u}}{{\partial x}} + 2y\cos 2u\dfrac{{\partial u}}{{\partial y}}\]
\[ \Rightarrow {x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}} = 2\cos 2u\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right)\]
Substitute \[\left( {x\dfrac{{\partial u}}{{\partial x}} + y\dfrac{{\partial u}}{{\partial y}}} \right) = \sin 2u\] in the above equation
\[ \Rightarrow {x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} + \sin 2u = 2\cos 2u\sin 2u\]
\[ \Rightarrow {x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} = 2\cos 2u\sin 2u - \sin 2u\]
Take common \[\sin 2u\] from right side expression
\[ \Rightarrow {x^2}\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} = \left( {2\cos 2u - 1} \right)\sin 2u\]

Hence option A is the correct option.

Note: The partial derivative of \[\dfrac{\partial f }{{\partial x}}\] with respect to \[y\] is \[\dfrac{\partial^{2} f }{{\partial y\partial x}}\]. The partial derivative of \[\dfrac{\partial f }{{\partial y}}\] with respect to \[x\] is \[\dfrac{\partial^{2} f }{{\partial x\partial y}}\]. The value of \[\dfrac{\partial^{2} f }{{\partial y\partial x}}\] is the same as \[\dfrac{\partial^{2} f }{{\partial x\partial y}}\]. Consider \[\dfrac{\partial y }{{\partial x}}=0\] and \[\dfrac{\partial x }{{\partial y}}=0\] for partial derivatives.