Question

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

Answer 1

Hint: 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 \[\dfrac{dy}{dx}\] to one side and solve it and again we will differentiate \[\dfrac{dy}{dx}\] again to obtain \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

Complete step-by-step solution:
Here, the given function is implicit function\[{{y}^{3}}+{{x}^{3}}-3axy=0\].
By differentiating it with respect to ’x’ we will have,
\[3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}-3a\left[ y\left( 1 \right)+x\left( \dfrac{dy}{dx} \right) \right]=0\] where, we got the equation using the multiplication rule, which is \[\dfrac{d}{dx}\left( uv \right)=\dfrac{du}{dx}\left( v \right)+\dfrac{dv}{dx}\left( u \right)\].
On further solving the equation, we get
\[3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}-3ay-3ax\dfrac{dy}{dx}=0\]
On simplifying, we get
\[{{x}^{2}}+{{y}^{2}}\dfrac{dy}{dx}-ay-ax\dfrac{dy}{dx}=0\]
Getting the \[\dfrac{dy}{dx}\] terms together, we get
\[\dfrac{dy}{dx}\left( {{y}^{2}}-ax \right)=ay-{{x}^{2}}\]
\[\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}\]
So, we have obtained the \[\dfrac{dy}{dx}\] , which is equals to \[\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}\].
Now, let us differentiate \[\dfrac{dy}{dx}\] again to find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\dfrac{d}{dx}\left( ay-{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\dfrac{d}{dx}\left( {{y}^{2}}-ax \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}\]
We obtained the above equation using division rule of differentiation, which is \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\], where $u=ay-{{x}^{2}}$ and $v={{y}^{2}}-ax$ .
On further solving we will have:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( a\dfrac{dy}{dx}-2x \right)-\left( ay-{{x}^{2}} \right)\left( 2y\dfrac{dy}{dx}-a \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}\].
Substituting \[\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}\]in the above equation, we have:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( a\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-2x \right)-\left( ay-{{x}^{2}} \right)\left( 2y\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-a \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}}\]
By expanding the above equation we have:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( {{a}^{2}}y-a{{x}^{2}}-2x{{y}^{2}}+2a{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\left( 2a{{y}^{2}}-2y{{x}^{2}}-a{{y}^{2}}+{{a}^{2}}x \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\]
On simplifying, we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left( {{a}^{2}}y+a{{x}^{2}}-2x{{y}^{2}}\right)-\left( ay-{{x}^{2}} \right)\left( a{{y}^{2}}-2y{{x}^{2}}+{{a}^{2}}x \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}{{y}^{3}}-2x{{y}^{4}}+a{{x}^{2}}{{y}^{2}}-{{a}^{3}}xy+2a{{x}^{2}}{{y}^{2}}-{{a}^{2}}{{x}^{3}}-\left( {{a}^{2}}{{y}^{3}}-2a{{x}^{2}}{{y}^{2}}+{{a}^{3}}xy-a{{x}^{2}}{{y}^{2}}+2{{x}^{4}}y-{{a}^{2}}{{x}^{3}} \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\]
Upon making the required cancellations, we have;
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy-2{{x}^{4}}y-2x{{y}^{4}}+6a{{x}^{2}}{{y}^{2}}}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy-2xy\left( {{x}^{3}}+{{y}^{3}}-3axy \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\].
Since \[{{y}^{3}}+{{x}^{3}}-3axy=0\], we have:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2{{a}^{3}}xy}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\]
So, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{a}^{3}}xy}{{{\left( ax-{{y}^{2}} \right)}^{3}}}\].
Hence \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{a}^{3}}xy}{{{\left( ax-{{y}^{2}} \right)}^{3}}}\] is the answer.

Note: While finding differentiation of an implicit function, firstly remember the product rule \[\dfrac{d}{dx}\left( uv \right)=\dfrac{du}{dx}\left( v \right)+\dfrac{dv}{dx}\left( u \right)\] and division rule \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\] are important while doing differentiation. It is suggested that first to simplify only \[\dfrac{dy}{dx}\]and then differentiate \[\dfrac{dy}{dx}\]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.