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Question: How do you determine \(\dfrac{dy}{dx}\) given \({{x}^{2}}y+y=3\) ?...

How do you determine dydx\dfrac{dy}{dx} given x2y+y=3{{x}^{2}}y+y=3 ?

Explanation

Solution

Problems of differentiation can be easily solved by expressing the equation as a function of xx first. Then differentiating the right-hand part which is totally a function of xx using a simple differentiation formula of chain rule of differentiation. We will reach the final result by simplifying the differentiated term at the end of the solution.

Complete step by step answer:
We could have started solving this problem by simply differentiating both the sides. But at the end we have to put the value of yy again. So, a better approach would be solving for yyfirst. In this way the problem becomes simplified already and can be easily solved by implicit differentiation.
That means the entire equation can be simplified by writing the xxrelated terms on one side and yyon the other. In this way we get yy as a function of xx as shown:
x2y+y=3{{x}^{2}}y+y=3
⇒y(x2+1)=3\Rightarrow y\left( {{x}^{2}}+1 \right)=3
⇒y=3x2+1\Rightarrow y=\dfrac{3}{{{x}^{2}}+1}
⇒y=3(x2+1)−1\Rightarrow y=3{{\left( {{x}^{2}}+1 \right)}^{-1}}
Now for differentiation we apply chain rule for the right-hand part. According to the chain rule of differentiation: ddxf(u(x))=f′(u(x))⋅u′(x)\dfrac{d}{dx}f\left( u\left( x \right) \right)=f'\left( u\left( x \right) \right)\cdot u'\left( x \right)
Here, the functions we have assumed are f(u(x))=3(x2+1)−1f\left( u\left( x \right) \right)=3{{\left( {{x}^{2}}+1 \right)}^{-1}} and u(x)=x2+1u\left( x \right)={{x}^{2}}+1 .
Taking the main equation y=3(x2+1)−1y=3{{\left( {{x}^{2}}+1 \right)}^{-1}} and differentiating on both the sides, we get
\dfrac{dy}{dx}=\dfrac{d\left\\{ 3{{\left( {{x}^{2}}+1 \right)}^{-1}} \right\\}}{dx}\cdot \dfrac{d\left( {{x}^{2}}+1 \right)}{dx}
\Rightarrow \dfrac{dy}{dx}=\left\\{ 3\left( -1 \right){{\left( {{x}^{2}}+1 \right)}^{-2}} \right\\}\cdot \left( 2x \right)
Further simplifying, we get
dydx=(−6x)(x2+1)−2\dfrac{dy}{dx}=\left( -6x \right){{\left( {{x}^{2}}+1 \right)}^{-2}}
Converting the term with negative power into reciprocal term of positive power, we have
dydx=−6x(x2+1)2\dfrac{dy}{dx}=-\dfrac{6x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}
Therefore, we can conclude to the simplified solution of the problem as dydx=−6x(x2+1)2\dfrac{dy}{dx}=-\dfrac{6x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} .

Note: While performing implicit differentiation we must be extra careful about assuming the functions, otherwise the problem can get complicated and further differentiation becomes hard. Also, while applying the chain rule we must take care about following the steps properly. A step jump during differentiating can cause error in the solution.