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Question: If \(u=\log \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\) , then prove that \(\left( {{x}^{2}}+{{y}^{2}}+{{...

If u=log⁡x2+y2+z2u=\log \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}} , then prove that (x2+y2+z2)(d2udx2+d2udy2+d2udz2)=1\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}\right)\left(\dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=1

Explanation

Solution

Hint: In the question, first work out on differentiation of the given function u two times with respective to x, y and z respectively. Second work out on the left hand side of the given expression and their simplification.

Complete step-by-step answer:
It is given that
u=log⁡x2+y2+z2u=\log \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}
u=log⁡(x2+y2+z2)12u=\log {{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}
By using the logarithmic rule, we get
u=12log⁡(x2+y2+z2)u=\dfrac{1}{2}\log \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)
Multiplying both sides by 2, we get
2u=log⁡(x2+y2+z2)................(1)2u=\log \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)................(1)
Differentiating with respect to x on both sides, we get
2dudx=1x2+y2+z2(2x)2\dfrac{du}{dx}=\dfrac{1}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}(2x)
Dividing both sides by 2, we get
dudx=xx2+y2+z2..............(2)\dfrac{du}{dx}=\dfrac{x}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}..............(2)
Differentiating the equation (2) with respect to x by using quotient rule (ddxuv)=v(dudx)−u(dvdx)v2\left( \dfrac{d}{dx}\dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}} , we get
d2udx2=(x2+y2+z2)(dxdx)−x(ddx(x2+y2+z2))(x2+y2+z2)2\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{dx}{dx} \right)-x\left( \dfrac{d}{dx}({{x}^{2}}+{{y}^{2}}+{{z}^{2}}) \right)}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}
Rearranging the terms, we get
d2udx2=(x2+y2+z2)−x(2x)(x2+y2+z2)2=x2+y2+z2−2x2(x2+y2+z2)2\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-x\left( 2x \right)}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}=\dfrac{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2{{x}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}
Finally we get
d2udx2=−x2+y2+z2(x2+y2+z2)2\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{-{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}
Similarly, differentiate the given function with respect to y and z , we get
d2udy2=x2−y2+z2(x2+y2+z2)2\dfrac{{{d}^{2}}u}{d{{y}^{2}}}=\dfrac{{{x}^{2}}-{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} and d2udz2=x2+y2−z2(x2+y2+z2)2\dfrac{{{d}^{2}}u}{d{{z}^{2}}}=\dfrac{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}
Let us consider the left hand side,
(x2+y2+z2)(d2udx2+d2udy2+d2udz2)=(x2+y2+z2)[−x2+y2+z2+x2−y2+z2+x2+y2−z2(x2+y2+z2)2]\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\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)\left[ \dfrac{-{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{x}^{2}}-{{y}^{2}}+{{z}^{2}}+{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]
Cancelling the like terms on the right side, we get
(x2+y2+z2)(d2udx2+d2udy2+d2udz2)=(x2+y2+z2)[x2+y2+z2(x2+y2+z2)2]\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\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)\left[ \dfrac{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]
Rearranging the terms, we get
(x2+y2+z2)(d2udx2+d2udy2+d2udz2)=[(x2+y2+z2)2(x2+y2+z2)2]\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\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{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]
Cancelling the terms on the right side, we get
(x2+y2+z2)(d2udx2+d2udy2+d2udz2)=1\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=1
This is a desired result.

Note:We might get confused about the difference between the ordinary differentiation and partial differentiation. In ordinary differentiation, we find derivatives with respect to one variable only, as function contains only one variable. Partial differentiation is used to differentiate mathematical functions having more than one variable in them.