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Question: How do you find \(\dfrac{dy}{dx}\) by implicit differentiation of \({{x}^{3}}+{{y}^{3}}=4xy+1\) and ...

How do you find dydx\dfrac{dy}{dx} by implicit differentiation of x3+y3=4xy+1{{x}^{3}}+{{y}^{3}}=4xy+1 and evaluate at point (2,1)\left( 2,1 \right)?

Explanation

Solution

To solve the given function by implicit differentiation we will differentiate the given equation with respect to x. We will use the chain rule to differentiate the left side of the equation and use the product rule to differentiate the right side of the equation. Then in the obtained value of dydx\dfrac{dy}{dx} we put the point (2,1)\left( 2,1 \right) to evaluate the value.

Complete step by step solution:
We have been given a function x3+y3=4xy+1{{x}^{3}}+{{y}^{3}}=4xy+1.
We have to solve the given function by implicit differentiation and evaluate dydx\dfrac{dy}{dx} at point (2,1)\left( 2,1 \right).
Now, to solve the given equation we will take the differential on both sides of the equation implicitly. Then we will get
⇒ddx(x3+y3)=ddx(4xy+1) ⇒ddxx3+ddxy3=4ddxxy+ddx1 \begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( {{x}^{3}}+{{y}^{3}} \right)=\dfrac{d}{dx}\left( 4xy+1 \right) \\\ & \Rightarrow \dfrac{d}{dx}{{x}^{3}}+\dfrac{d}{dx}{{y}^{3}}=4\dfrac{d}{dx}xy+\dfrac{d}{dx}1 \\\ \end{aligned}
Now, we know that ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} and ddx(xy)=xddxy+yddxx\dfrac{d}{dx}\left( xy \right)=x\dfrac{d}{dx}y+y\dfrac{d}{dx}x
So by applying the formula to the above obtained equation we will get
⇒3x2+3y2dydx=4(xddxy+yddxx)+0\Rightarrow 3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}=4\left( x\dfrac{d}{dx}y+y\dfrac{d}{dx}x \right)+0
Now, simplifying the above obtained equation we will get
⇒3x2+3y2dydx=4(xdydx+y) ⇒3x2+3y2dydx=4xdydx+4y \begin{aligned} & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}=4\left( x\dfrac{dy}{dx}+y \right) \\\ & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}\dfrac{dy}{dx}=4x\dfrac{dy}{dx}+4y \\\ \end{aligned}
Now, taking the common terms out we will get
⇒3y2dydx−4xdydx=4y−3x2 ⇒dydx(3y2−4x)=4y−3x2 \begin{aligned} & \Rightarrow 3{{y}^{2}}\dfrac{dy}{dx}-4x\dfrac{dy}{dx}=4y-3{{x}^{2}} \\\ & \Rightarrow \dfrac{dy}{dx}\left( 3{{y}^{2}}-4x \right)=4y-3{{x}^{2}} \\\ \end{aligned}
Now, rearranging the terms to get the value of dydx\dfrac{dy}{dx} we will get
⇒dydx=4y−3x23y2−4x\Rightarrow \dfrac{dy}{dx}=\dfrac{4y-3{{x}^{2}}}{3{{y}^{2}}-4x}
Now, to evaluate dydx\dfrac{dy}{dx} at point (2,1)\left( 2,1 \right), we will substitute the value x=2x=2 and y=1y=1 in the above obtained equation. Then we will get
⇒dydx=4×1−3(22)3×12−4×2\Rightarrow \dfrac{dy}{dx}=\dfrac{4\times 1-3\left( {{2}^{2}} \right)}{3\times {{1}^{2}}-4\times 2}
Now, simplifying the above obtained equation we will get

& \Rightarrow \dfrac{dy}{dx}=\dfrac{4-12}{3-8} \\\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{-8}{-5} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{8}{5} \\\ \end{aligned}$$ Hence we get the value of $$\dfrac{dy}{dx}=\dfrac{8}{5}$$ for the given points. **Note:** As there is a difference between chain rule and product rule. Product rule is applied when we have the product of two functions. We have to apply chain rule when we have a composite function like ‘function of function’. They are not separate functions. So be careful while applying these rules.