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Question: How do you find \(\dfrac{dy}{dx}\) by implicit differentiation of \({{\left( x+y \right)}^{3}}={{x}^...

How do you find dydx\dfrac{dy}{dx} by implicit differentiation of (x+y)3=x3+y3{{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}} and evaluate at point (−1,1)\left( -1,1 \right)?

Explanation

Solution

As we know that in implicit function the function has both terms x and y. So to solve the given function by implicit differentiation we will differentiate the given equation with respect to x. Then in the obtained value of dydx\dfrac{dy}{dx} we put the point (−1,1)\left( -1,1 \right) to evaluate the value.

Complete step by step solution:
We have been given a function (x+y)3=x3+y3{{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}.
We have to solve the given function by implicit differentiation and evaluate dydx\dfrac{dy}{dx} at point (−1,1)\left( -1,1 \right).
Now, to solve the given equation we will take the differential on both sides of the equation implicitly. Then we will get
⇒ddx(x+y)3=ddx(x3+y3)\Rightarrow \dfrac{d}{dx}{{\left( x+y \right)}^{3}}=\dfrac{d}{dx}\left( {{x}^{3}}+{{y}^{3}} \right)
Now, as the function on the RHS has both x and y terms so we need to apply chain rule. Then we will get
⇒3(x+y)2ddx(x+y)=ddxx3+ddxy3\Rightarrow 3{{\left( x+y \right)}^{2}}\dfrac{d}{dx}\left( x+y \right)=\dfrac{d}{dx}{{x}^{3}}+\dfrac{d}{dx}{{y}^{3}}
Now, we know that ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}
So by applying the formula to the above obtained equation we will get
⇒3(x+y)2(1+dydx)=3x2+3y2dydx\Rightarrow 3{{\left( x+y \right)}^{2}}\left( 1+\dfrac{dy}{dx} \right)=3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}
Now, simplifying the above obtained equation we will get
⇒3(x+y)2+3(x+y)2dydx=3x2+3y2dydx ⇒3(x+y)2dydx−3y2dydx=3x2−3(x+y)2 \begin{aligned} & \Rightarrow 3{{\left( x+y \right)}^{2}}+3{{\left( x+y \right)}^{2}}\dfrac{dy}{dx}=3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx} \\\ & \Rightarrow 3{{\left( x+y \right)}^{2}}\dfrac{dy}{dx}-3{{y}^{2}}\dfrac{dy}{dx}=3{{x}^{2}}-3{{\left( x+y \right)}^{2}} \\\ \end{aligned}
Now, taking the common terms out we will get
⇒3dydx(x2+y2+2xy−y2)=3x2−3(x2+y2+2xy)\Rightarrow 3\dfrac{dy}{dx}\left( {{x}^{2}}+{{y}^{2}}+2xy-{{y}^{2}} \right)=3{{x}^{2}}-3\left( {{x}^{2}}+{{y}^{2}}+2xy \right)
Now, simplifying the above obtained equation we will get
⇒3dydx(x2+2xy)=−3(y2+2xy)\Rightarrow 3\dfrac{dy}{dx}\left( {{x}^{2}}+2xy \right)=-3\left( {{y}^{2}}+2xy \right)
Now, rearranging the terms to get the value of dydx\dfrac{dy}{dx} we will get

& \Rightarrow \dfrac{dy}{dx}=-\dfrac{3\left( {{y}^{2}}+2xy \right)}{3\left( {{x}^{2}}+2xy \right)} \\\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( {{y}^{2}}+2xy \right)}{\left( {{x}^{2}}+2xy \right)} \\\ \end{aligned}$$ Now, to evaluate $$\dfrac{dy}{dx}$$ at point $\left( -1,1 \right)$, we will substitute the value $x=-1$ and $y=1$ in the above obtained equation. Then we will get $$\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( {{1}^{2}}+2\left( -1 \right)1 \right)}{\left( {{\left( -1 \right)}^{2}}+2\left( -1 \right)1 \right)}$$ Now, simplifying the above obtained equation we will get $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( 1-2 \right)}{\left( 1-2 \right)} \\\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{-1}{-1} \\\ & \Rightarrow \dfrac{dy}{dx}=-1 \\\ \end{aligned}$$ Hence we get the value of $$\dfrac{dy}{dx}=-1$$ for the given points. **Note:** In implicit differentiation we have to treat y same as the x and when we differentiate y we have to multiply it by $\dfrac{dy}{dx}$. Always bring all the other terms except $$\dfrac{dy}{dx}$$ to one side to easily get the answer. Students must have knowledge of basic differentiation rules to solve such types of questions.