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Question: If \({{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}\) , show that \(\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\...

If xmyn=(x+y)m+n{{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}} , show that d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}} = 0.

Explanation

Solution

In this problem, we are given an equation which is to be used to prove that the double derivative with respect to y will be equal to zero. We will be using some properties of logarithms and rules for differentiation. First we have to solve for a single derivative. Then find a double derivative.

Complete step by step answer:
Let’s solve the problem now.
Let’s have a look on some logarithm rules:
Product rule: log⁡bMN=log⁡bM+log⁡bN{{\log }_{b}}MN={{\log }_{b}}M+{{\log }_{b}}N
Quotient rule: log⁡MN=log⁡bM−log⁡bN\log \dfrac{M}{N}={{\log }_{b}}M-{{\log }_{b}}N
Power rule: log⁡bMp=plog⁡bM{{\log }_{_{b}}}{{M}^{p}}=p{{\log }_{b}}M
Now, we have to discuss some of the important rules of differentiation also. Suppose we have two functions say f and g are given. Then we will see how the rules can be applied on them.
Power rule: ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}
Sum rule: ddx(f±g)=f′±g′\dfrac{d}{dx}\left( f\pm g \right)=f'\pm g'
Product rule: ddx(fg)=fg′+f′g\dfrac{d}{dx}\left( fg \right)=fg'+f'g
Quotient rule: ddx(fg)=f′g−fg′g2\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{f'g-fg'}{{{g}^{2}}}
Write the equation given:
⇒xmyn=(x+y)m+n\Rightarrow {{x}^{m}}{{y}^{n}}={{\left( x+y \right)}^{m+n}}
Take log on both the sides:
⇒log⁡xmyn=log⁡(x+y)m+n\Rightarrow \log {{x}^{m}}{{y}^{n}}=\log {{\left( x+y \right)}^{m+n}}
By using power rule and product rule for logarithms i.e.
Product rule: log⁡bMN=log⁡bM+log⁡bN{{\log }_{b}}MN={{\log }_{b}}M+{{\log }_{b}}N
Power rule: log⁡bMp=plog⁡bM{{\log }_{_{b}}}{{M}^{p}}=p{{\log }_{b}}M
Apply these two rules in above equation:
⇒log⁡xm+log⁡yn=(m+n)log⁡(x+y)\Rightarrow \log {{x}^{m}}+\log {{y}^{n}}=\left( m+n \right)\log \left( x+y \right)
Differentiate both side with respect to ‘x’, we get:
⇒m.1x+n1ydydx=(m+n).1(x+y)(1+dydx)\Rightarrow m.\dfrac{1}{x}+n\dfrac{1}{y}\dfrac{dy}{dx}=\left( m+n \right).\dfrac{1}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right)
Simplify terms:
⇒mx+nydydx=(m+n)(x+y)(1+dydx)\Rightarrow \dfrac{m}{x}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{\left( m+n \right)}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right)
Now, open the bracket and multiply the terms:
⇒mx+nydydx=m+nx+y+(m+nx+y)dydx\Rightarrow \dfrac{m}{x}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{m+n}{x+y}+\left( \dfrac{m+n}{x+y} \right)\dfrac{dy}{dx}
Take dydx\dfrac{dy}{dx} containing terms on one side and rest on other side:
⇒nydydx−(m+nx+y)dydx=m+nx+y−mx\Rightarrow \dfrac{n}{y}\dfrac{dy}{dx}-\left( \dfrac{m+n}{x+y} \right)\dfrac{dy}{dx}=\dfrac{m+n}{x+y}-\dfrac{m}{x}
Take dydx\dfrac{dy}{dx} common from left hand side:
⇒(ny−m+nx+y)dydx=m+nx+y−mx\Rightarrow \left( \dfrac{n}{y}-\dfrac{m+n}{x+y} \right)\dfrac{dy}{dx}=\dfrac{m+n}{x+y}-\dfrac{m}{x}
Now, by cross multiplying and making the denominator common, solve the equation:
⇒(n(x+y)−y(m+n)y(x+y))dydx=x(m+n)−m(x+y)x(x+y)\Rightarrow \left( \dfrac{n\left( x+y \right)-y\left( m+n \right)}{y\left( x+y \right)} \right)\dfrac{dy}{dx}=\dfrac{x\left( m+n \right)-m\left( x+y \right)}{x\left( x+y \right)}
On simplifying the equation, we will get:
⇒(nx+ny−my−nyy(x+y))dydx=xm+xn−mx−myx(x+y)\Rightarrow \left( \dfrac{nx+ny-my-ny}{y\left( x+y \right)} \right)\dfrac{dy}{dx}=\dfrac{xm+xn-mx-my}{x\left( x+y \right)}
Cancel the like terms:
⇒(nx−myy(x+y))dydx=xn−myx(x+y)\Rightarrow \left( \dfrac{nx-my}{y\left( x+y \right)} \right)\dfrac{dy}{dx}=\dfrac{xn-my}{x\left( x+y \right)}
We can also write this equation as:
⇒1(x+y).(nx−myy)dydx=1(x+y).xn−myx\Rightarrow \dfrac{1}{\left( x+y \right)}.\left( \dfrac{nx-my}{y} \right)\dfrac{dy}{dx}=\dfrac{1}{\left( x+y \right)}.\dfrac{xn-my}{x}
Cancel 1x+y\dfrac{1}{x+y} on both the sides, we get:
⇒(nx−myy)dydx=xn−myx\Rightarrow \left( \dfrac{nx-my}{y} \right)\dfrac{dy}{dx}=\dfrac{xn-my}{x}
Expression nx - my will be cancelled from both the sides:
⇒dydx=yx.....(i)\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}.....(i)
So we obtained a single derivative for now. But we have to find a double derivative as well. So again differentiate with respect to x, we will get:
⇒d2ydx2=ddx(yx)\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{y}{x} \right)
Apply quotient rule: ddx(fg)=f′g−fg′g2\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{f'g-fg'}{{{g}^{2}}}, we get:
⇒d2ydx2=x.dydx−y.1x2⇔x.dydx−yx2\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x.\dfrac{dy}{dx}-y.1}{{{x}^{2}}}\Leftrightarrow \dfrac{x.\dfrac{dy}{dx}-y}{{{x}^{2}}}
If we write this expression more clearly, it will look like:
⇒d2ydx2=1x.dydx−yx2\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}.\dfrac{dy}{dx}-\dfrac{y}{{{x}^{2}}}
Use in equation(i) in above equation:
⇒d2ydx2=1x.yx−yx2\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}.\dfrac{y}{x}-\dfrac{y}{{{x}^{2}}}
Multiply the terms:
⇒d2ydx2=yx2−yx2\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y}{{{x}^{2}}}-\dfrac{y}{{{x}^{2}}}
On solving further, we get:
⇒d2ydx2=0\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0
Hence, it’s proved.

Note: Take care of the integers as well as algebraic terms while solving differentiation questions. Do remember that ddxlog⁡x=1x\dfrac{d}{dx}\log x=\dfrac{1}{x}. This derivative is used in between the answer which is left unnoticed. Perform each step carefully. Don’t miss the steps in order to complete the answer fast because there are many calculations.