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Question: How do you use implicit differentiation to find \(\dfrac{dy}{dx}\) given \(16{{x}^{2}}+25{{y}^{2}}=4...

How do you use implicit differentiation to find dydx\dfrac{dy}{dx} given 16x2+25y2=40016{{x}^{2}}+25{{y}^{2}}=400?

Explanation

Solution

To solve the given question we will use the concept of implicit differentiation. We will differentiate both sides of the given equation with respect to x. In order to differentiate the equation we will use the power rule of the differentiation.

Complete step-by-step answer:
We have been given an equation 16x2+25y2=40016{{x}^{2}}+25{{y}^{2}}=400.
We have to differentiate the given equation using implicit differentiation.
Now, differentiating the given equation with respect to x we will get
ddx(16x2)+ddx(25y2)=ddx400\Rightarrow \dfrac{d}{dx}\left( 16{{x}^{2}} \right)+\dfrac{d}{dx}\left( 25{{y}^{2}} \right)=\dfrac{d}{dx}400
Now, we know that ddxxn=nxn1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}
Now, applying the differentiation rules to the above obtained equation we will get
16×2x+25×2ydydx=0\Rightarrow 16\times 2x+25\times 2y\dfrac{dy}{dx}=0
Now, simplifying the above obtained equation we will get
32x+50ydydx=0\Rightarrow 32x+50y\dfrac{dy}{dx}=0
Now, to find the value of dydx\dfrac{dy}{dx} we need to rearrange the terms in the above obtained equation. Then we will get
50ydydx=32x dydx=32x50y \begin{aligned} & \Rightarrow 50y\dfrac{dy}{dx}=-32x \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-32x}{50y} \\\ \end{aligned}
Now, simplifying the above obtained equation we will get
dydx=16x25y\Rightarrow \dfrac{dy}{dx}=\dfrac{-16x}{25y}
Hence above is the required solution of the given equation.

Note: The possibility of mistake is that if students consider y as a constant and solve accordingly they will get the incorrect solution. Alternatively we can simplify the equation and convert it into the function of x alone and then differentiate the equation. The point to be remembered while solving the implicit differentiation is that we need to add dydx\dfrac{dy}{dx} every time we find a y term in the equation.