Answer
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Hint: To differentiate an implicit equation with respect to one of the variables, we differentiate both the sides of the equation with respect to that variable. Hence, differentiate both the sides with respect to x. You may also have to use the product rule of differentiation.
Complete step-by-step solution:
Let us first understand what is meant by implicit equations.
In simple terms, an implicit equation is an equation that has more than one variable and not in the form $y=f(x)$ where f is some function of x.
To differentiate an implicit equation with respect to one of the variables, we differentiate both the sides of the equation with respect to that variable.
Here, the given equation is $1-xy=x-y$ …. (i).
Let us assume that the independent variable of the equation is x and the variable y is the dependent variable and depends on x.
Therefore, let us differentiate equation (i) with respect to x.
Then,
$\Rightarrow \dfrac{d}{dx}\left( 1-xy \right)=\dfrac{d}{dx}\left( x-y \right)$
Using the associative property of differentiation we can write the above equation as
$\Rightarrow \dfrac{d}{dx}\left( 1 \right)-\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( x \right)-\dfrac{d}{dx}\left( y \right)$ …. (ii)
Differentiation of a constant is always equal to zero.
Therefore, $\dfrac{d}{dx}\left( 1 \right)=0$
And $\dfrac{d}{dx}\left( x \right)$
The derivative $\dfrac{d}{dx}\left( xy \right)$ can be simplified using product rule.
According to the product rule of differentiation, $\dfrac{d}{dx}(xy)=\dfrac{d}{dx}(x).y+x.\dfrac{d}{dx}(y)$
This further implies that $\dfrac{d}{dx}(xy)=(1)y+x\dfrac{dy}{dx}$
Which means that $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$
Now, substitute all the given values in equation (ii).
$\Rightarrow 0-\left( y+x\dfrac{dy}{dx} \right)=1-\dfrac{dy}{dx}$
Then,
$\Rightarrow \dfrac{dy}{dx}-x\dfrac{dy}{dx}=1+y$
$\Rightarrow (1-x)\dfrac{dy}{dx}=1+y$
This means that $\dfrac{dy}{dx}=\dfrac{1+y}{1-x}$ …. (iii)
Hence, we calculated the derivative of y with respect to x by differentiating implicitly.
Note: When we have an equation in the form $y=f(x)$ where f is some function of x, this equation is called an explicit function. In this, we find the derivative by just differentiating the right hand side of the equation (i.e. f(x)). Sometimes, the question may demand to find the derivative in the term of x only. Then you can substitute the value of y in equation (iii) from the first equation given in the question.
Complete step-by-step solution:
Let us first understand what is meant by implicit equations.
In simple terms, an implicit equation is an equation that has more than one variable and not in the form $y=f(x)$ where f is some function of x.
To differentiate an implicit equation with respect to one of the variables, we differentiate both the sides of the equation with respect to that variable.
Here, the given equation is $1-xy=x-y$ …. (i).
Let us assume that the independent variable of the equation is x and the variable y is the dependent variable and depends on x.
Therefore, let us differentiate equation (i) with respect to x.
Then,
$\Rightarrow \dfrac{d}{dx}\left( 1-xy \right)=\dfrac{d}{dx}\left( x-y \right)$
Using the associative property of differentiation we can write the above equation as
$\Rightarrow \dfrac{d}{dx}\left( 1 \right)-\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( x \right)-\dfrac{d}{dx}\left( y \right)$ …. (ii)
Differentiation of a constant is always equal to zero.
Therefore, $\dfrac{d}{dx}\left( 1 \right)=0$
And $\dfrac{d}{dx}\left( x \right)$
The derivative $\dfrac{d}{dx}\left( xy \right)$ can be simplified using product rule.
According to the product rule of differentiation, $\dfrac{d}{dx}(xy)=\dfrac{d}{dx}(x).y+x.\dfrac{d}{dx}(y)$
This further implies that $\dfrac{d}{dx}(xy)=(1)y+x\dfrac{dy}{dx}$
Which means that $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$
Now, substitute all the given values in equation (ii).
$\Rightarrow 0-\left( y+x\dfrac{dy}{dx} \right)=1-\dfrac{dy}{dx}$
Then,
$\Rightarrow \dfrac{dy}{dx}-x\dfrac{dy}{dx}=1+y$
$\Rightarrow (1-x)\dfrac{dy}{dx}=1+y$
This means that $\dfrac{dy}{dx}=\dfrac{1+y}{1-x}$ …. (iii)
Hence, we calculated the derivative of y with respect to x by differentiating implicitly.
Note: When we have an equation in the form $y=f(x)$ where f is some function of x, this equation is called an explicit function. In this, we find the derivative by just differentiating the right hand side of the equation (i.e. f(x)). Sometimes, the question may demand to find the derivative in the term of x only. Then you can substitute the value of y in equation (iii) from the first equation given in the question.
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