Question

If \[u = {\left( {{x^2} + {y^2} + {z^2}} \right)^{\dfrac{1}{2}}}\] , then prove that \[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = 2\].

Answer 1

Hint: First, differentiate the given equation with respect to the variable \[x\]. Again, differentiate the first derivative with respect to the variable \[x\] by using the quotient rule of differentiation. Follow these steps to calculate the second derivative of the given equation with respect to \[y\], and \[z\]. In the end, substitute the values of all three-second derivatives on the left-hand side of the given equation to reach the required answer.

Formula used:
 \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
\[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right) = \dfrac{1}{{2\sqrt {{x^n}} }}\dfrac{d}{{dx}}\left( {{x^n}} \right)\]
Quotient rule: \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]

Complete step by step solution:
The given equation is \[u = \sqrt {{x^2} + {y^2} + {z^2}} \].
To prove: \[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = 2\]

Let’s differentiate the given equation with respect to the variable \[x\].
Apply the differential formula \[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right) = \dfrac{1}{{2\sqrt {{x^n}} }}\dfrac{d}{{dx}}\left( {{x^n}} \right)\]
\[\dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}\dfrac{d}{{dx}}\left( {{x^2} + {y^2} + {z^2}} \right)\]
\[ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {2x} \right)\]
\[ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\]
Again, differentiate the above differential equation with respect to the variable \[x\].
\[\dfrac{d}{{dx}}\left( {\dfrac{{du}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)\]
Apply the quotient rule of differentiation.
\[\dfrac{{{d^2}u}}{{d{x^2}}} = \dfrac{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)\dfrac{d}{{dx}}\left( x \right) - x\dfrac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^2}}}\]
\[ \Rightarrow \dfrac{{{d^2}u}}{{d{x^2}}} = \dfrac{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right) - x\dfrac{1}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {2x} \right)}}{{\left( {{x^2} + {y^2} + {z^2}} \right)}}\]
\[ \Rightarrow \dfrac{{{d^2}u}}{{d{x^2}}} = \dfrac{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right) - \dfrac{{{x^2}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}}}{{\left( {{x^2} + {y^2} + {z^2}} \right)}}\]
Simplify the above equation.
\[\dfrac{{{d^2}u}}{{d{x^2}}} = \dfrac{{{x^2} + {y^2} + {z^2} - {x^2}}}{{\left( {{x^2} + {y^2} + {z^2}} \right)\sqrt {{x^2} + {y^2} + {z^2}} }}\]
\[ \Rightarrow \dfrac{{{d^2}u}}{{d{x^2}}} = \dfrac{{{y^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}\] \[.....\left( 1 \right)\]

Similarly, double differentiate the given equation with respect to the variables \[y\] and \[z\].
We get,
\[\dfrac{{{d^2}u}}{{d{y^2}}} = \dfrac{{{x^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}\] \[.....\left( 2 \right)\]
 and \[\dfrac{{{d^2}u}}{{d{z^2}}} = \dfrac{{{x^2} + {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}\] \[.....\left( 3 \right)\]

Now substitute the equations \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\] in the given expression \[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right)\].
\[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = u\left[ {\dfrac{{{y^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{{{x^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{{{x^2} + {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}} \right]\]
Substitute \[u = {\left( {{x^2} + {y^2} + {z^2}} \right)^{\dfrac{1}{2}}}\] in the above equation.
\[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = {\left( {{x^2} + {y^2} + {z^2}} \right)^{\dfrac{1}{2}}}\left[ {\dfrac{{{y^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{{{x^2} + {z^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{{{x^2} + {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}} \right]\]
Simplify the right-hand side of the above equation.
\[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = {\left( {{x^2} + {y^2} + {z^2}} \right)^{\dfrac{1}{2}}}\left[ {\dfrac{{{y^2} + {z^2} + {x^2} + {z^2} + {x^2} + {y^2}}}{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\dfrac{3}{2}}}}}} \right]\]
Cancel out the common terms from the numerator and denominator.
\[u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = \left[ {\dfrac{{2\left( {{x^2} + {y^2} + {z^2}} \right)}}{{\left( {{x^2} + {y^2} + {z^2}} \right)}}} \right]\]
\[ \Rightarrow u\left( {\dfrac{{{d^2}u}}{{d{x^2}}} + \dfrac{{{d^2}u}}{{d{y^2}}} + \dfrac{{{d^2}u}}{{d{z^2}}}} \right) = 2\]
Hence, proved.

Note: Students often get confused about the differential formula of \[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right)\] that whether \[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right) = \dfrac{1}{{2\sqrt {{x^n}} }}\dfrac{d}{{dx}}\left( {{x^n}} \right)\] or \[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right) = \dfrac{1}{{2\sqrt {{x^n}} }}\]. But the correct formula is \[\dfrac{d}{{dx}}\left( {\sqrt {{x^n}} } \right) = \dfrac{1}{{2\sqrt {{x^n}} }}\dfrac{d}{{dx}}\left( {{x^n}} \right)\].