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Question: If u = f(r), where \[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\] then \[\left( \dfrac{{{\partial }^{2}}u}{\parti...

If u = f(r), where r2=x2+y2{{r}^{2}}={{x}^{2}}+{{y}^{2}} then (∂2u∂x2+∂2u∂y2)=\left( \dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}} \right)=
(a) f′(r)\left( \text{a} \right)\text{ }{{f}^{'}}\left( r \right)
(b) f′′(r)+f′(r)\left( \text{b} \right)\text{ }{{f}^{''}}\left( r \right)+{{f}^{'}}\left( r \right)
(c) f′′(r)+1rf′(r)\left( \text{c} \right)\text{ }{{f}^{''}}\left( r \right)+\dfrac{1}{r}{{f}^{'}}\left( r \right)
(d) f′(r)+rf′(r)\left( \text{d} \right)\text{ }{{f}^{'}}\left( r \right)+r{{f}^{'}}\left( r \right)

Explanation

Solution

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 ∂u∂x\dfrac{\partial u}{\partial x} by differentiating u = f(r) and we will use ∂u∂x=∂u∂r×∂r∂x.\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\times \dfrac{\partial r}{\partial x}. After getting the value of ∂u∂x,\dfrac{\partial u}{\partial x}, we will differentiate it once again and we will find ∂2u∂x2\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}} in the terms of x, r, ∂u∂r\dfrac{\partial u}{\partial r} and ∂2u∂r2.\dfrac{{{\partial }^{2}}u}{\partial {{r}^{2}}}. Similarly, we will find the value of ∂2u∂y2\dfrac{{{\partial }^{2}}u}{\partial {{y}^{2}}} and then we will add ∂2u∂x2\dfrac{{{\partial }^{2}}u}{\partial {{x}^{2}}} and ∂2u∂y2\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(r).....(i)u=f\left( r \right).....\left( i \right)
r2=x2+y2.....(ii){{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,
∂u∂x=∂f(r)∂x......(iii)\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,
⇒∂u∂x=∂f(r)∂x×∂r∂r\Rightarrow \dfrac{\partial u}{\partial x}=\dfrac{\partial f\left( r \right)}{\partial x}\times \dfrac{\partial r}{\partial r}
⇒∂u∂x=∂f(r)∂r×∂r∂x\Rightarrow \dfrac{\partial u}{\partial x}=\dfrac{\partial f\left( r \right)}{\partial r}\times \dfrac{\partial r}{\partial x}
We will write ∂f(r)∂r\dfrac{\partial f\left( r \right)}{\partial r} as f’(r) in the above equation. Thus, we will get,
⇒∂u∂x=f′(r)×∂r∂x......(iv)\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,
∂∂x(r2)=∂∂x(x2)+∂∂x(y2)\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 x2{{x}^{2}} is 2x and y2{{y}^{2}} is zero because y is constant and differentiation of constant is zero. Thus, we will get,
∂∂x(r2)=2x+0\dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)=2x+0
⇒∂∂x(r2)=2x\Rightarrow \dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)=2x
Multiplying numerator and denominator by ∂r\partial r on LHS, we get,
⇒∂∂x(r2)×∂r∂r=2x\Rightarrow \dfrac{\partial }{\partial x}\left( {{r}^{2}} \right)\times \dfrac{\partial r}{\partial r}=2x
⇒∂∂r(r2)×∂r∂x=2x\Rightarrow \dfrac{\partial }{\partial r}\left( {{r}^{2}} \right)\times \dfrac{\partial r}{\partial x}=2x
⇒2r×∂r∂x=2x\Rightarrow 2r\times \dfrac{\partial r}{\partial x}=2x
⇒r∂r∂x=x\Rightarrow \dfrac{r\partial r}{\partial x}=x
⇒∂r∂x=xr.....(v)\Rightarrow \dfrac{\partial r}{\partial x}=\dfrac{x}{r}.....\left( v \right)
Now, we will put the value of ∂r∂x\dfrac{\partial r}{\partial x} from (v) to (iv). Thus, we will get,
⇒∂u∂x=f′(r).xr.....(vi)\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,
∂∂x(∂u∂x)=∂∂x[f′(r).xr]\dfrac{\partial }{\partial x}\left( \dfrac{\partial u}{\partial x} \right)=\dfrac{\partial }{\partial x}\left[ {{f}^{'}}\left( r \right).\dfrac{x}{r} \right]
⇒∂2u∂x2=∂∂x[f′(r).xr]\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,
dda[P(a).Q(a)]=Q(a).dda[P(a)]+P(a).dda[Q(a)]\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,
⇒∂2u∂x2=xr[∂∂x(f′(r))]+f′(r)[∂∂x(xr)]......(vii)\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 ∂∂x(f′(r)).\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,
ddx[A(B(x))]=A′(B(x))×B′(x)\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,
∂∂x(f′(r))=f′′(r).∂r∂x\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 ∂r∂x\dfrac{\partial r}{\partial x} from (v) to the above equation. Thus, we will get,
∂∂x(f′(r))=f′′(r).xr......(viii)\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 ∂∂x(xr).\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,
ddx(AB)=B.dA−A.dBB2\dfrac{d}{dx}\left( \dfrac{A}{B} \right)=\dfrac{B.dA-A.dB}{{{B}^{2}}}
Thus, we will get,
∂∂x(xr)=r.(∂x∂x)−x.(∂r∂x)r2\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}}}
⇒∂∂x(xr)=r−x(∂r∂x)r2\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 ∂r∂u\dfrac{\partial r}{\partial u} from (v) to the above equation. Thus, we will get,
⇒∂∂x(xr)=r−x(xr)r2=r2−x2r3......(ix)\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,
∂2u∂x2=xr[f′′(r).xr]+f′(r)[r2−x2r3]\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]
⇒∂2u∂x2=f′′(r)x2r2+f′(r).(r2−x2)r3......(x)\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,
∂2u∂y2=f′′(r).y2r2+f′(r).(r2−y2)r3......(xi)\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,
∂2u∂x2+∂2u∂y2=f′′(r).x2r2+f′(r).(r2−x2)r3+f′′(r).y2r2+f′(r).(r2−y2)r3\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}}}
⇒∂2u∂x2+∂2u∂y2=f′′(r)[x2r2+y2r2]+f′(r)r3[r2−x2+r2−y2]\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]
⇒∂2u∂x2+∂2u∂y2=f′′(r)[r2r2]+f′(r)r3[2r2−r2]\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]
⇒∂2u∂x2+∂2u∂y2=f′′(r)+f′(r)r3.r2\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}}
⇒∂2u∂x2+∂2u∂y2=f′′(r)+f′(r)r\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 x1,{{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.