Question

How do you find $\dfrac{dy}{dx}$ by implicit differentiation given $x{{y}^{2}}-3{{x}^{2}}y+x=1$?

Answer 1

This question belongs to the topic of implicit differentiation which belongs to the topic of calculus. In this question, we are going to use a differentiation formula that is a product rule. In solving this question, we will first differentiate the whole equation (which is given in the question) with respect to x. After doing further differentiation, we will use the product rule and solve the further differentiation. After doing some processes, we will get the answer.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find $\dfrac{dy}{dx}$ from the equation $x{{y}^{2}}-3{{x}^{2}}y+x=1$ by implicit differentiation. So, we have to find $\dfrac{dy}{dx}$ using the implicit differentiation.
The equation which we have to differentiate is
$x{{y}^{2}}-3{{x}^{2}}y+x=1$
Now, we will differentiate both sides of the above equation with respect to x. We can write
$\Rightarrow \dfrac{d}{dx}\left( x{{y}^{2}}-3{{x}^{2}}y+x \right)=\dfrac{d}{dx}\left( 1 \right)$
As we know that the differentiation of any constant is 0, so we can write the above differentiation as
$\Rightarrow \dfrac{d}{dx}\left( x{{y}^{2}}-3{{x}^{2}}y+x \right)=0$
Now, using the formula of addition rule of differentiation that is $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}\left( b \right)$, we can write the above differentiation as
$\Rightarrow \dfrac{d}{dx}\left( x{{y}^{2}} \right)-\dfrac{d}{dx}\left( 3{{x}^{2}}y \right)+\dfrac{d}{dx}\left( x \right)=0$
Now, using the product rule of differentiation that is $\dfrac{d}{dx}\left( a\times b \right)=b\dfrac{d}{dx}\left( a \right)+a\dfrac{d}{dx}\left( b \right)$, we can write
$\Rightarrow {{y}^{2}}\dfrac{d}{dx}\left( x \right)+x\dfrac{d}{dx}\left( {{y}^{2}} \right)-y\dfrac{d}{dx}\left( 3{{x}^{2}} \right)-3{{x}^{2}}\dfrac{d}{dx}\left( y \right)+\dfrac{d}{dx}\left( x \right)=0$
Now, using the formula $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$, we can write the above differentiation as
$\Rightarrow {{y}^{2}}\dfrac{dx}{dx}+x\left( 2y\dfrac{d}{dx}y \right)-y\left( 3\times 2x \right)-3{{x}^{2}}\left( \dfrac{d}{dx}y \right)+\dfrac{dx}{dx}=0$
The above equation can also be written as
$\Rightarrow {{y}^{2}}\times 1+2xy\dfrac{dy}{dx}-6xy-3{{x}^{2}}\dfrac{dy}{dx}+1=0$
Now, taking common $\dfrac{dy}{dx}$ we can write
$\Rightarrow {{y}^{2}}-6xy+\left( 2xy-3{{x}^{2}} \right)\dfrac{dy}{dx}+1=0$
The above equation can also be written as
$\Rightarrow \left( 2xy-3{{x}^{2}} \right)\dfrac{dy}{dx}=6xy-{{y}^{2}}-1$
The above equation can also be written as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 6xy-{{y}^{2}}-1 \right)}{\left( 2xy-3{{x}^{2}} \right)}$
Hence, we have found the value of $\dfrac{dy}{dx}$ from the equation $x{{y}^{2}}-3{{x}^{2}}y+x=1$. The value of $\dfrac{dy}{dx}$ is $\dfrac{\left( 6xy-{{y}^{2}}-1 \right)}{\left( 2xy-3{{x}^{2}} \right)}$.

Note:
We should have a better knowledge in the topic of implicit differentiation. We should remember the following formulas for solving this type of question easily:
Product rule of differentiation: $\dfrac{d}{dx}\left( a\times b \right)=b\dfrac{d}{dx}\left( a \right)+a\dfrac{d}{dx}\left( b \right)$
Addition rule of differentiation: $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}\left( b \right)$
Power rule of differentiation: $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
We should know how to find implicit differentiation to solve this type of question easily.