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Question: If we have an expression as \[{{y}^{3}}+{{x}^{3}}-3axy=0\], then find the second derivative \[\dfrac...

If we have an expression as y3+x3−3axy=0{{y}^{3}}+{{x}^{3}}-3axy=0, then find the second derivative d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.

Explanation

Solution

As the given function is an implicit function, what we will do is we will differentiate the given function with respect to x and then we will collect all dydx\dfrac{dy}{dx} to one side and solve it and again we will differentiate dydx\dfrac{dy}{dx} again to obtain d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.

Complete step-by-step solution:
Here, the given function is implicit functiony3+x3−3axy=0{{y}^{3}}+{{x}^{3}}-3axy=0.
By differentiating it with respect to ’x’ we will have,
3x2+3y2dydx−3a[y(1)+x(dydx)]=03{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}-3a\left[ y\left( 1 \right)+x\left( \dfrac{dy}{dx} \right) \right]=0 where, we got the equation using the multiplication rule, which is ddx(uv)=dudx(v)+dvdx(u)\dfrac{d}{dx}\left( uv \right)=\dfrac{du}{dx}\left( v \right)+\dfrac{dv}{dx}\left( u \right).
On further solving the equation, we get
3x2+3y2dydx−3ay−3axdydx=03{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}-3ay-3ax\dfrac{dy}{dx}=0
On simplifying, we get
x2+y2dydx−ay−axdydx=0{{x}^{2}}+{{y}^{2}}\dfrac{dy}{dx}-ay-ax\dfrac{dy}{dx}=0
Getting the dydx\dfrac{dy}{dx} terms together, we get
dydx(y2−ax)=ay−x2\dfrac{dy}{dx}\left( {{y}^{2}}-ax \right)=ay-{{x}^{2}}
dydx=ay−x2y2−ax\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}
So, we have obtained the dydx\dfrac{dy}{dx} , which is equals to dydx=ay−x2y2−ax\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}.
Now, let us differentiate dydx\dfrac{dy}{dx} again to find d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.
d2ydx2=(y2−ax)ddx(ay−x2)−(ay−x2)ddx(y2−ax)(y2−ax)2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\dfrac{d}{dx}\left( ay-{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\dfrac{d}{dx}\left( {{y}^{2}}-ax \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}
We obtained the above equation using division rule of differentiation, which is ddx(uv)=vdudx−udvdxv2\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}, where u=ay−x2u=ay-{{x}^{2}} and v=y2−axv={{y}^{2}}-ax .
On further solving we will have:
d2ydx2=(y2−ax)(adydx−2x)−(ay−x2)(2ydydx−a)(y2−ax)2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( a\dfrac{dy}{dx}-2x \right)-\left( ay-{{x}^{2}} \right)\left( 2y\dfrac{dy}{dx}-a \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}.
Substituting dydx=ay−x2y2−ax\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}in the above equation, we have:
d2ydx2=(y2−ax)(a(ay−x2y2−ax)−2x)−(ay−x2)(2y(ay−x2y2−ax)−a)(y2−ax)2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( a\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-2x \right)-\left( ay-{{x}^{2}} \right)\left( 2y\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-a \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}
By expanding the above equation we have:
d2ydx2=(y2−ax)(a2y−ax2−2xy2+2ax2)−(ay−x2)(2ay2−2yx2−ay2+a2x)(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( {{a}^{2}}y-a{{x}^{2}}-2x{{y}^{2}}+2a{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\left( 2a{{y}^{2}}-2y{{x}^{2}}-a{{y}^{2}}+{{a}^{2}}x \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}
On simplifying, we get
d2ydx2=(y2−ax)(a2y+ax2−2xy2)−(ay−x2)(ay2−2yx2+a2x)(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( {{a}^{2}}y+a{{x}^{2}}-2x{{y}^{2}}\right)-\left( ay-{{x}^{2}} \right)\left( a{{y}^{2}}-2y{{x}^{2}}+{{a}^{2}}x \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}
d2ydx2=a2y3−2xy4+ax2y2−a3xy+2ax2y2−a2x3−(a2y3−2ax2y2+a3xy−ax2y2+2x4y−a2x3)(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}{{y}^{3}}-2x{{y}^{4}}+a{{x}^{2}}{{y}^{2}}-{{a}^{3}}xy+2a{{x}^{2}}{{y}^{2}}-{{a}^{2}}{{x}^{3}}-\left( {{a}^{2}}{{y}^{3}}-2a{{x}^{2}}{{y}^{2}}+{{a}^{3}}xy-a{{x}^{2}}{{y}^{2}}+2{{x}^{4}}y-{{a}^{2}}{{x}^{3}} \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}
Upon making the required cancellations, we have;
d2ydx2=−2a3xy−2x4y−2xy4+6ax2y2(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy-2{{x}^{4}}y-2x{{y}^{4}}+6a{{x}^{2}}{{y}^{2}}}{{{\left( {{y}^{2}}-ax \right)}^{3}}}
d2ydx2=−2a3xy−2xy(x3+y3−3axy)(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy-2xy\left( {{x}^{3}}+{{y}^{3}}-3axy \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}.
Since y3+x3−3axy=0{{y}^{3}}+{{x}^{3}}-3axy=0, we have:
d2ydx2=−2a3xy(y2−ax)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy}{{{\left( {{y}^{2}}-ax \right)}^{3}}}
So, d2ydx2=2a3xy(ax−y2)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{a}^{3}}xy}{{{\left( ax-{{y}^{2}} \right)}^{3}}}.
Hence d2ydx2=2a3xy(ax−y2)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{a}^{3}}xy}{{{\left( ax-{{y}^{2}} \right)}^{3}}} is the answer.

Note: While finding differentiation of an implicit function, firstly remember the product rule ddx(uv)=dudx(v)+dvdx(u)\dfrac{d}{dx}\left( uv \right)=\dfrac{du}{dx}\left( v \right)+\dfrac{dv}{dx}\left( u \right) and division rule ddx(uv)=vdudx−udvdxv2\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} are important while doing differentiation. It is suggested that first to simplify only dydx\dfrac{dy}{dx}and then differentiate dydx\dfrac{dy}{dx}using division rule as this will help you in solving question with less error. The question is easy but there are chances of making calculation errors, so try to avoid making calculation mistakes.