Question

If u = f(r), where \[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\] then \[\left( \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}} \right)=\]
\[\left( \text{a} \right)\text{ }{{f}^{'}}\left( r \right)\]
\[\left( \text{b} \right)\text{ }{{f}^{''}}\left( r \right)+{{f}^{'}}\left( r \right)\]
\[\left( \text{c} \right)\text{ }{{f}^{''}}\left( r \right)+\dfrac{1}{r}{{f}^{'}}\left( r \right)\]
\[\left( \text{d} \right)\text{ }{{f}^{'}}\left( r \right)+r{{f}^{'}}\left( r \right)\]

Answer 1

Hint: To solve the given question, we will first find out about the operation \[\partial \] and what it does when it is applied on some variable or function. Then, we will find the value of \[\dfrac{\partial u}{\partial x}\] by differentiating u = f(r) and we will use \[\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\times \dfrac{\partial r}{\partial x}.\] After getting the value of \[\dfrac{\partial u}{\partial x},\] we will differentiate it once again and we will find \[\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}\] in the terms of x, r, \[\dfrac{\partial u}{\partial r}\] and \[\dfrac{{{\partial }^{2}}u}{\partial {{r}^{2}}}.\] Similarly, we will find the value of \[\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}\] and then we will add \[\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}\] and \[\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}\] to get the final answer.

Complete step-by-step answer:
Before we solve the given question, we must know what is \[\partial \] operator. \[\partial \] operator is the operator for partial differentiation of a variable with respect to another variable where we treat the other variables as constant. Now, we are given that,
\[u=f\left( r \right).....\left( i \right)\]
\[{{r}^{2}}={{x}^{2}}+{{y}^{2}}.....\left( ii \right)\]
Now, we will differentiate (i) with respect to x on both sides. Thus, we will get,
\[\dfrac{\partial u}{\partial x}=\dfrac{\partial f\left( r \right)}{\partial x}......\left( iii \right)\]
We can write (iii) also in the following manner,
\[\Rightarrow \dfrac{\partial u}{\partial x}=\dfrac{\partial f\left( r \right)}{\partial x}\times \dfrac{\partial r}{\partial r}\]
\[\Rightarrow \dfrac{\partial u}{\partial x}=\dfrac{\partial f\left( r \right)}{\partial r}\times \dfrac{\partial r}{\partial x}\]
We will write \[\dfrac{\partial f\left( r \right)}{\partial r}\] as f’(r) in the above equation. Thus, we will get,
\[\Rightarrow \dfrac{\partial u}{\partial x}={{f}^{'}}\left( r \right)\times \dfrac{\partial r}{\partial x}......\left( iv \right)\]
Now, we will partially differentiate (ii) with respect to x, treating y as constant. Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)=\dfrac{\partial }{\partial x}\left( {{x}^{2}} \right)+\dfrac{\partial }{\partial x}\left( {{y}^{2}} \right)\]
Now, the differentiation of \[{{x}^{2}}\] is 2x and \[{{y}^{2}}\] is zero because y is constant and differentiation of constant is zero. Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)=2x+0\]
\[\Rightarrow \dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)=2x\]
Multiplying numerator and denominator by \[\partial r\] on LHS, we get,
\[\Rightarrow \dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)\times \dfrac{\partial r}{\partial r}=2x\]
\[\Rightarrow \dfrac{\partial }{\partial r}\left( {{r}^{2}} \right)\times \dfrac{\partial r}{\partial x}=2x\]
\[\Rightarrow 2r\times \dfrac{\partial r}{\partial x}=2x\]
\[\Rightarrow \dfrac{r\partial r}{\partial x}=x\]
\[\Rightarrow \dfrac{\partial r}{\partial x}=\dfrac{x}{r}.....\left( v \right)\]
Now, we will put the value of \[\dfrac{\partial r}{\partial x}\] from (v) to (iv). Thus, we will get,
\[\Rightarrow \dfrac{\partial u}{\partial x}={{f}^{'}}\left( r \right).\dfrac{x}{r}.....\left( vi \right)\]
Now, we will differentiate (vi) with respect to x again. Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( \dfrac{\partial u}{\partial x} \right)=\dfrac{\partial }{\partial x}\left[ {{f}^{'}}\left( r \right).\dfrac{x}{r} \right]\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\dfrac{\partial }{\partial x}\left[ {{f}^{'}}\left( r \right).\dfrac{x}{r} \right]\]
Now, we will use the product rule of differentiation in the above equation. This rule says that,
\[\dfrac{d}{da}\left[ P\left( a \right).Q\left( a \right) \right]=Q\left( a \right).\dfrac{d}{da}\left[ P\left( a \right) \right]+P\left( a \right).\dfrac{d}{da}\left[ Q\left( a \right) \right]\]
Thus, we will get,
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\dfrac{x}{r}\left[ \dfrac{\partial }{\partial x}\left( {{f}^{'}}\left( r \right) \right) \right]+{{f}^{'}}\left( r \right)\left[ \dfrac{\partial }{\partial x}\left( \dfrac{x}{r} \right) \right]......\left( vii \right)\]
Now, we will find the value of \[\dfrac{\partial }{\partial x}\left( {{f}^{'}}\left( r \right) \right).\] For this, we will use the chain rule of differentiation. According to this rule, we have,
\[\dfrac{d}{dx}\left[ A\left( B\left( x \right) \right) \right]={{A}^{'}}\left( B\left( x \right) \right)\times {{B}^{'}}\left( x \right)\]
Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( {{f}^{'}}\left( r \right) \right)={{f}^{''}}\left( r \right).\dfrac{\partial r}{\partial x}\]
Now, we will put the value of \[\dfrac{\partial r}{\partial x}\] from (v) to the above equation. Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( {{f}^{'}}\left( r \right) \right)={{f}^{''}}\left( r \right).\dfrac{x}{r}......\left( viii \right)\]
Now, we will find the value of \[\dfrac{\partial }{\partial x}\left( \dfrac{x}{r} \right).\] For this, we will use the division rule of differentiation. According to this rule, we have,
\[\dfrac{d}{dx}\left( \dfrac{A}{B} \right)=\dfrac{B.dA-A.dB}{{{B}^{2}}}\]
Thus, we will get,
\[\dfrac{\partial }{\partial x}\left( \dfrac{x}{r} \right)=\dfrac{r.\left( \dfrac{\partial x}{\partial x} \right)-x.\left( \dfrac{\partial r}{\partial x} \right)}{{{r}^{2}}}\]
\[\Rightarrow \dfrac{\partial }{\partial x}\left( \dfrac{x}{r} \right)=\dfrac{r-x\left( \dfrac{\partial r}{\partial x} \right)}{{{r}^{2}}}\]
Now, we will put the value of \[\dfrac{\partial r}{\partial u}\] from (v) to the above equation. Thus, we will get,
\[\Rightarrow \dfrac{\partial }{\partial x}\left( \dfrac{x}{r} \right)=\dfrac{r-x\left( \dfrac{x}{r} \right)}{{{r}^{2}}}=\dfrac{{{r}^{2}}-{{x}^{2}}}{{{r}^{3}}}......\left( ix \right)\]
From (vii), (viii) and (ix), we have,
\[\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\dfrac{x}{r}\left[ {{f}^{''}}\left( r \right).\dfrac{x}{r} \right]+{{f}^{'}}\left( r \right)\left[ \dfrac{{{r}^{2}}-{{x}^{2}}}{{{r}^{3}}} \right]\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}={{f}^{''}}\left( r \right)\dfrac{{{x}^{2}}}{{{r}^{2}}}+{{f}^{'}}\left( r \right).\dfrac{\left( {{r}^{2}}-{{x}^{2}} \right)}{{{r}^{3}}}......\left( x \right)\]
Similarly,
\[\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right).\dfrac{{{y}^{2}}}{{{r}^{2}}}+{{f}^{'}}\left( r \right).\dfrac{\left( {{r}^{2}}-{{y}^{2}} \right)}{{{r}^{3}}}......\left( xi \right)\]
On adding (x) and (xi), we get,
\[\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right).\dfrac{{{x}^{2}}}{{{r}^{2}}}+{{f}^{'}}\left( r \right).\dfrac{\left( {{r}^{2}}-{{x}^{2}} \right)}{{{r}^{3}}}+{{f}^{''}}\left( r \right).\dfrac{{{y}^{2}}}{{{r}^{2}}}+{{f}^{'}}\left( r \right).\dfrac{\left( {{r}^{2}}-{{y}^{2}} \right)}{{{r}^{3}}}\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right)\left[ \dfrac{{{x}^{2}}}{{{r}^{2}}}+\dfrac{{{y}^{2}}}{{{r}^{2}}} \right]+\dfrac{{{f}^{'}}\left( r \right)}{{{r}^{3}}}\left[ {{r}^{2}}-{{x}^{2}}+{{r}^{2}}-{{y}^{2}} \right]\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right)\left[ \dfrac{{{r}^{2}}}{{{r}^{2}}} \right]+\dfrac{{{f}^{'}}\left( r \right)}{{{r}^{3}}}\left[ 2{{r}^{2}}-{{r}^{2}} \right]\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right)+\dfrac{{{f}^{'}}\left( r \right)}{{{r}^{3}}}.{{r}^{2}}\]
\[\Rightarrow \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}}={{f}^{''}}\left( r \right)+\dfrac{{{f}^{'}}\left( r \right)}{r}\]
Hence, the option (c) is the right answer.

Note: While solving the question, we have assumed that the function U is differentiable two times with respect to \[{{x}_{1}},\] r and y, i.e. the double differentiation of U with respect to x and y exists. Another thing to remember is that while differentiating partially, we take all the remaining variables as constant as long as there is no relation given between them.