Question

If \[z = \tan\left( {y - ax} \right) + {\left( {y - ax} \right)^{\dfrac{3}{2}}}\] , then show that \[\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = {a^2}\dfrac{{{\partial ^2}z}}{{\partial {y^2}}}\].

Answer 1

Hint: First, differentiate the given equation partially with respect to \[x\]. Again, differentiate the given equation partially with respect to \[x\] and calculate the second-order derivative. Follow these steps to calculate the second-order derivative of the given equation with respect to \[y\]. In the end, compare the derivatives to reach the required answer.

Formula used:
\[\dfrac{\partial }{{\partial x}}\left( {\tan x} \right) = \sec^{2}x\]
\[\dfrac{\partial }{{\partial x}}\left( {\sec x} \right) = \sec x \tan x\]
\[\dfrac{\partial }{{\partial x}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Product rule of partial differentiation: \[\dfrac{\partial }{{\partial x}}\left( {uv} \right) = u\dfrac{{\partial v}}{{\partial x}} + v\dfrac{{\partial u}}{{\partial x}}\]

Complete step by step solution:
The given trigonometric equation is \[z = \tan\left( {y - ax} \right) + {\left( {y - ax} \right)^{\dfrac{3}{2}}}\].
To Prove: \[\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = {a^2}\dfrac{{{\partial ^2}u}}{{\partial {y^2}}}\]
Let’s differentiate the above equation partially with respect to the variable \[x\].
\[\dfrac{\partial }{{\partial x}}\left( z \right) = \dfrac{\partial }{{\partial x}}\left( {\tan\left( {y - ax} \right) + {{\left( {y - ax} \right)}^{\dfrac{3}{2}}}} \right)\]
Apply the partial differentiation formulas \[\dfrac{\partial }{{\partial x}}\left( {\tan x} \right) = \sec^{2}x\], \[\dfrac{\partial }{{\partial x}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and chain rule for \[\dfrac{\partial }{{\partial x}}\left( {\tan\left( {y - ax} \right)} \right)\].
\[\dfrac{{\partial z}}{{\partial x}} = \left( {\sec^{2}\left( {y - ax} \right)\dfrac{\partial }{{\partial x}}\left( {y - ax} \right)} \right) + \left( {\dfrac{3}{2}{{\left( {y - ax} \right)}^{\dfrac{1}{2}}}\dfrac{\partial }{{\partial x}}\left( {y - ax} \right)} \right)\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = - a\sec^{2}\left( {y - ax} \right) - \dfrac{{3a}}{2}{\left( {y - ax} \right)^{\dfrac{1}{2}}}\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial x}} = - \left( {a\sec^{2}\left( {y - ax} \right) + \dfrac{{3a}}{2}{{\left( {y - ax} \right)}^{\dfrac{1}{2}}}} \right)\]
Again, differentiate the above differential equation partially with respect to the variable \[x\].
\[\dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial z}}{{\partial x}}} \right) = - \dfrac{\partial }{{\partial x}}\left( {a\sec^{2}\left( {y - ax} \right) + \dfrac{{3a}}{2}{{\left( {y - ax} \right)}^{\dfrac{1}{2}}}} \right)\]
Apply the chain rule for \[\dfrac{\partial }{{\partial x}}\left( {a\sec^{2}\left( {y - ax} \right)} \right)\].
\[\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = - \left( {2a\sec\left( {y - ax} \right)\dfrac{\partial }{{\partial x}}\left( {\sec\left( {y - ax} \right)} \right) + \dfrac{{3a}}{2}\left( {\dfrac{1}{{2\sqrt {y - ax} }}} \right)\dfrac{\partial }{{\partial x}}\left( {y - ax} \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = - \left( {2a\sec\left( {y - ax} \right)\sec\left( {y - ax} \right)\tan\left( {y - ax} \right)\dfrac{\partial }{{\partial x}}\left( {y - ax} \right) + \dfrac{{3a}}{{4\sqrt {y - ax} }}\left( { - a} \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = - \left( {2a\sec^{2}\left( {y - ax} \right)\tan\left( {y - ax} \right)\left( { - a} \right) - \dfrac{{3{a^2}}}{{4\sqrt {y - ax} }}} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = - \left( { - 2{a^2}\sec^{2}\left( {y - ax} \right)\tan\left( {y - ax} \right) - \dfrac{{3{a^2}}}{{4\sqrt {y - ax} }}} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = {a^2}\left( {2\sec^{2}\left( {y - ax} \right)\tan\left( {y - ax} \right) + \dfrac{3}{{4\sqrt {y - ax} }}} \right)\] \[.....\left( 1 \right)\]
Now differentiate the given equation partially with respect to the variable \[y\].
\[\dfrac{\partial }{{\partial y}}\left( z \right) = \dfrac{\partial }{{\partial y}}\left( {\tan\left( {y - ax} \right) + {{\left( {y - ax} \right)}^{\dfrac{3}{2}}}} \right)\]
Apply the partial differentiation formulas \[\dfrac{\partial }{{\partial x}}\left( {\tan x} \right) = \sec^{2}x\], \[\dfrac{\partial }{{\partial x}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and chain rule for \[\dfrac{\partial }{{\partial y}}\left( {\tan\left( {y - ax} \right)} \right)\].
\[\dfrac{{\partial z}}{{\partial y}} = \left( {\sec^{2}\left( {y - ax} \right)\dfrac{\partial }{{\partial y}}\left( {y - ax} \right)} \right) + \left( {\dfrac{3}{2}{{\left( {y - ax} \right)}^{\dfrac{1}{2}}}\dfrac{\partial }{{\partial y}}\left( {y - ax} \right)} \right)\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \sec^{2}\left( {y - ax} \right)\left( 1 \right) + \dfrac{3}{2}{\left( {y - ax} \right)^{\dfrac{1}{2}}}\left( 1 \right)\]
\[ \Rightarrow \dfrac{{\partial z}}{{\partial y}} = \sec^{2}\left( {y - ax} \right) + \dfrac{3}{2}{\left( {y - ax} \right)^{\dfrac{1}{2}}}\]
Again, differentiate the above differential equation partially with respect to the variable \[y\].
\[\dfrac{\partial }{{\partial y}}\left( {\dfrac{{\partial z}}{{\partial y}}} \right) = \dfrac{\partial }{{\partial y}}\left( {\sec^{2}\left( {y - ax} \right) + \dfrac{3}{2}{{\left( {y - ax} \right)}^{\dfrac{1}{2}}}} \right)\]
Apply the chain rule for \[\dfrac{\partial }{{\partial y}}\left( {\sec^{2}\left( {y - ax} \right)} \right)\].
\[\dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = \left( {2\sec\left( {y - ax} \right)\dfrac{\partial }{{\partial y}}\left( {\sec\left( {y - ax} \right)} \right) + \dfrac{3}{2}\left( {\dfrac{1}{{2\sqrt {y - ax} }}} \right)\dfrac{\partial }{{\partial y}}\left( {y - ax} \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = \left( {2\sec\left( {y - ax} \right)\sec\left( {y - ax} \right)\tan\left( {y - ax} \right)\dfrac{\partial }{{\partial y}}\left( {y - ax} \right) + \dfrac{3}{2}\left( {\dfrac{1}{{2\sqrt {y - ax} }}} \right)\left( 1 \right)} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = \left( {2\sec^{2}\left( {y - ax} \right)\tan\left( {y - ax} \right)\left( 1 \right) + \dfrac{3}{{4\sqrt {y - ax} }}} \right)\]
\[ \Rightarrow \dfrac{{{\partial ^2}z}}{{\partial {y^2}}} = 2\sec^{2}\left( {y - ax} \right)\tan\left( {y - ax} \right) + \dfrac{3}{{4\sqrt {y - ax} }}\] \[.....\left( 2 \right)\]

Now compare the equations \[\left( 1 \right)\], and \[\left( 2 \right)\].
We get,
\[\dfrac{{{\partial ^2}z}}{{\partial {x^2}}} = {a^2}\dfrac{{{\partial ^2}z}}{{\partial {y^2}}}\]
Hence, proved.

Note: Students often get confused and directly apply the product rule without using the chain rule of differentiation.
The chain rule is used to take derivatives of composites of functions and this happens by chaining together their derivatives.
Chain rule: If \[F\left( x \right) = f\left( {g\left( x \right)} \right)\], then \[F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\]