Find the derivative of the following:
\[{{y}^{2}}-3xy+{{x}^{2}}=x\]
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Hint: If u and v are functions in terms of x then their differentiation with respect to x is given by\[\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}\]and this formula is called as the product rule. Use the product rule to simplify the given equation and calculate the derivative of the equation.
Complete step-by-step answer:
We know the differentiation with respect to x for two functions u and v in terms of x is given by \[\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}\].
Applying the above mentioned formula to simplify the given equation.
First apply derivative on both sides of the equation with respect to x then we will get
\[\dfrac{d}{dx}\left( {{y}^{2}} \right)-3\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}\left( x \right)\]
Using the formula for derivative of \[{{u}^{n}}\] in terms of u is given by \[\dfrac{d}{du}\left( {{u}^{n}} \right)=n{{u}^{n-1}}\] where n is any integer and u is any variable we get,
We know the differentiation of x with respect to x is 1.
\[2y\dfrac{dy}{dx}-3\left( x\times \dfrac{dy}{dx}+y \right)+2x=1\]
Simplifying the equation we get,
\[2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}+3y+2x=1\]
Subtracting with 3y and 2x on both sides we get,
\[2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}=1-3y-2x\]
Taking \[\dfrac{dy}{dx}\] as common in the LHS we get,
\[\dfrac{dy}{dx}\left( 2y-3x \right)=1-3y-2x\]
On dividing with (2y - 3x) on both sides we get,
\[\dfrac{dy}{dx}=\dfrac{1-3y-2x}{2y-3x}\]
Hence, we get the derivative of the equation \[{{y}^{2}}-3xy+{{x}^{2}}=x\] as \[\dfrac{dy}{dx}=\dfrac{1-3y-2x}{2y-3x}\].
Note: The formula for derivative of \[{{u}^{n}}\] in terms of u is given by \[\dfrac{d}{du}\left( {{u}^{n}} \right)=n{{u}^{n-1}}\] where n is any integer and u is any variable. The derivative of x component with respect to x is 1 and derivative of y component with respect to x is \[\dfrac{dy}{dx}\]. Using previously mentioned rules carefully complete the basic mathematical operations like addition, subtraction to calculate the final answer.
Complete step-by-step answer:
We know the differentiation with respect to x for two functions u and v in terms of x is given by \[\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}\].
Applying the above mentioned formula to simplify the given equation.
First apply derivative on both sides of the equation with respect to x then we will get
\[\dfrac{d}{dx}\left( {{y}^{2}} \right)-3\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}\left( x \right)\]
Using the formula for derivative of \[{{u}^{n}}\] in terms of u is given by \[\dfrac{d}{du}\left( {{u}^{n}} \right)=n{{u}^{n-1}}\] where n is any integer and u is any variable we get,
We know the differentiation of x with respect to x is 1.
\[2y\dfrac{dy}{dx}-3\left( x\times \dfrac{dy}{dx}+y \right)+2x=1\]
Simplifying the equation we get,
\[2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}+3y+2x=1\]
Subtracting with 3y and 2x on both sides we get,
\[2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}=1-3y-2x\]
Taking \[\dfrac{dy}{dx}\] as common in the LHS we get,
\[\dfrac{dy}{dx}\left( 2y-3x \right)=1-3y-2x\]
On dividing with (2y - 3x) on both sides we get,
\[\dfrac{dy}{dx}=\dfrac{1-3y-2x}{2y-3x}\]
Hence, we get the derivative of the equation \[{{y}^{2}}-3xy+{{x}^{2}}=x\] as \[\dfrac{dy}{dx}=\dfrac{1-3y-2x}{2y-3x}\].
Note: The formula for derivative of \[{{u}^{n}}\] in terms of u is given by \[\dfrac{d}{du}\left( {{u}^{n}} \right)=n{{u}^{n-1}}\] where n is any integer and u is any variable. The derivative of x component with respect to x is 1 and derivative of y component with respect to x is \[\dfrac{dy}{dx}\]. Using previously mentioned rules carefully complete the basic mathematical operations like addition, subtraction to calculate the final answer.