Answer

Verified

381k+ views

**Hint:**For doing the implicit differentiation for the given equation ${{x}^{2}}{{y}^{2}}+xy=2$ we first have to choose our dependent and the independent variables. As per the usual convention, we choose the independent variable as $x$ and the dependent variable as $y$. Then we have to differentiate both sides of the given equation with respect to the independent variable $x$. From the differentiated expression, we have to separate $\dfrac{dy}{dx}$ in terms of the variables \[x\] and \[y\] to obtain the final derivative.

**Complete step-by-step answer:**

The equation given in the question is written as

${{x}^{2}}{{y}^{2}}+xy=2$

The implicit differentiation means finding out the derivative of the dependent variable with respect to the independent variable without expressing it explicitly in the form of the independent variable. So first we have to choose our dependent and the independent variables from the given equation. Let us choose $y$ as the dependent variable and $x$ as the independent variable. So we have to find out the derivative of $y$ with respect to $x$. For this, we differentiate both sides of the above equation with respect to $x$ to get

\[\begin{align}

& \Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}}+xy \right)}{dx}=\dfrac{d\left( 2 \right)}{dx} \\

& \Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}} \right)}{dx}+\dfrac{d\left( xy \right)}{dx}=\dfrac{d\left( 2 \right)}{dx} \\

\end{align}\]

We know that the differentiation of a constant is equal to zero. So the RHS of the above equation becomes zero and we get

$\Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}} \right)}{dx}+\dfrac{d\left( xy \right)}{dx}=0$

Now, from the product rule of differentiation we know that $\dfrac{d\left( fg \right)}{dx}=f\dfrac{d\left( g \right)}{dx}+g\dfrac{d\left( f \right)}{dx}$. So the above equation can be written as

\[\begin{align}

& \Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dx}+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y\dfrac{dx}{dx}=0 \\

& \Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dx}+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y=0 \\

\end{align}\]

From the chain rule of differentiation, we write the first term of the above equation as

\[\Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dy}\left( \dfrac{dy}{dx} \right)+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y=0\]

Now, we know that the differentiation of the function ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So the above equation is written as

\[\begin{align}

& \Rightarrow {{x}^{2}}\left( 2{{y}^{2-1}} \right)\dfrac{dy}{dx}+{{y}^{2}}\left( 2{{x}^{2-1}} \right)+x\dfrac{dy}{dx}+y=0 \\

& \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+2x{{y}^{2}}+x\dfrac{dy}{dx}+y=0 \\

\end{align}\]

Writing the terms containing $\dfrac{dy}{dx}$ together, we have

\[\Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}+2x{{y}^{2}}+y=0\]

Subtracting \[2x{{y}^{2}}+y\] from both the sides, we have

\[\begin{align}

& \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}+2x{{y}^{2}}+y-\left( 2x{{y}^{2}}+y \right)=0-\left( 2x{{y}^{2}}+y \right) \\

& \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}=-\left( 2x{{y}^{2}}+y \right) \\

\end{align}\]

Taking \[\dfrac{dy}{dx}\] common

\[\Rightarrow \left( 2{{x}^{2}}y+x \right)\dfrac{dy}{dx}=-\left( 2x{{y}^{2}}+y \right)\]

Finally, dividing both sides by \[\left( 2{{x}^{2}}y+x \right)\] we get

\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( 2x{{y}^{2}}+y \right)}{\left( 2{{x}^{2}}y+x \right)}\]

Now, taking out $y$ and $x$ common from the numerator and denominator respectively, we get

\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{y\left( 2xy+1 \right)}{x\left( 2xy+1 \right)}\]

Finally cancelling \[\left( 2xy+1 \right)\], we get

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x}$

Hence, this is the required implicit differentiation.

**Note:**It is not compulsory to choose $y$ as the dependent variable and $x$ as the independent variable. We can do the reverse too. Also, try to simplify the final expression for the derivative as much is possible. For example, in this case do not conclude \[\dfrac{dy}{dx}=-\dfrac{\left( 2x{{y}^{2}}+y \right)}{\left( 2{{x}^{2}}y+x \right)}\] as the final derivative, as the factor \[\left( 2xy+1 \right)\] is still common to the numerator and the denominator which can be cancelled.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

How many crores make 10 million class 7 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE