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Question: How do you differentiate \(2xy+{{y}^{2}}=x+y\)?...

How do you differentiate 2xy+y2=x+y2xy+{{y}^{2}}=x+y?

Explanation

Solution

Derivative is the rate of change of a function. To find the derivative of a function, we have to differentiate it with respect to xx. Finding out the derivative of a function implies that we are finding out the slope of the function.dydx\dfrac{dy}{dx} tells us about the slope of a curve. In trigonometric terms, tan⁡θ\tan \theta , where θ\theta is the inclination of a line with the positive xx-axis measured in anti-clockwise direction , tells us about the slope of a line.

Complete step by step solution:
Here we have 2xy+y2=x+y2xy+{{y}^{2}}=x+y.
Now let us differentiate this with respect to xx.
Before doing that, we should know the following :
⇒d(xn)dx=nxn−1\Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} .
And also we need to make use of the uvuv rule of differentiation. Whenever we have two functions namely, h(x),g(x)h\left( x \right),g\left( x \right) in this form h(x)g(x)h\left( x \right)g\left( x \right) , then we differentiate such a function using this rule. It states the following :
⇒ddx(h(x)g(x))=g(x)ddx(h(x))+h(x)ddx(g(x))\Rightarrow \dfrac{d}{dx}\left( h\left( x \right)g\left( x \right) \right)=g\left( x \right)\dfrac{d}{dx}\left( h\left( x \right) \right)+h\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right) .
Let us use this piece of information while differentiating.
Now, let us differentiate.
Upon differentiating, we get the following :
⇒2xy+y2=x+y ⇒2y+2xdydx+2ydydx=1+dydx \begin{aligned} & \Rightarrow 2xy+{{y}^{2}}=x+y \\\ & \Rightarrow 2y+2x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\\ \end{aligned}
Let us gather all the dydx\dfrac{dy}{dx} onto one side of the equation.
Upon doing so, we get the following :
⇒2xy+y2=x+y ⇒2y+2xdydx+2ydydx=1+dydx ⇒dydx(2x+2y−1)+2y=1 \begin{aligned} & \Rightarrow 2xy+{{y}^{2}}=x+y \\\ & \Rightarrow 2y+2x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}\left( 2x+2y-1 \right)+2y=1 \\\ \end{aligned}
Let us send the 2y2yonto the right hand side of the equation from the left hand side.
Upon doing so, we get the following :
⇒2xy+y2=x+y ⇒2y+2xdydx+2ydydx=1+dydx ⇒dydx(2x+2y−1)+2y=1 ⇒dydx(2x+2y−1)=1−2y \begin{aligned} & \Rightarrow 2xy+{{y}^{2}}=x+y \\\ & \Rightarrow 2y+2x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}\left( 2x+2y-1 \right)+2y=1 \\\ & \Rightarrow \dfrac{dy}{dx}\left( 2x+2y-1 \right)=1-2y \\\ \end{aligned}
Let us bring the 2x+2y−12x+2y-1 onto the left hand side of the equation.
Upon doing so, we get the following :
⇒dydx=1−2y2x+2y−1\Rightarrow \dfrac{dy}{dx}=\dfrac{1-2y}{2x+2y-1}.

∴\therefore Hence , the derivative of the function 2xy+y2=x+y2xy+{{y}^{2}}=x+y is 1−2y2x+2y−1\dfrac{1-2y}{2x+2y-1}.

Note: We have to be careful while applying the formula. We should carefully compare and find out what h(x)&g(x)h\left( x \right)\And g\left( x \right) are . It is advisable to learn the derivatives of all the standard functions in order to solve the question quickly. We should also learn all the rules of differentiation.