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(b) Use implicit differentiation to find \frac{d y}{d x} for the curve \[ 2 x^{4}+x^{2} y-x y^{3}=3 \]

User Laffoyb
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Given curve is\[2x^4+x^2y-xy^3=3\] We need to find \[\frac {d y}{d x}] by implicit differentiation.

To differentiate this equation, we follow these steps :

Step 1: Differentiate each term of the equation with respect to x. (treat y as a function of x)

Step 2: Simplify the equation by collecting the like terms of \frac{d y}{d x}\] and then solve it for \[\frac{d y}{d x}\].

We start by differentiating each term with respect to x.

So, Step 1:\[2x^4+x^2y-xy^3=3\] Differentiate each term with respect to x.\[\frac{d}{d x}(2x^4)+\frac{d y}{dx}(x^2y)-\frac{d y}{d x}(xy^3)=\frac{d y}{d x}(3)\]

Simplify the terms,\8x^3+2xy+x^2\frac{d y}{d x}-y^3-x^3\frac{d y}{d x}3xy^2=0\]

Grouping the terms of \[\frac{d y}{dx}\] on one side and constant terms on the other side,\[x^2\frac{d y}{d x}-x^3\frac{d y}{d x}3xy^2=-(8x^3+2xy-y^3)\]\[\frac{d y}{d x}(x^2-x^3.3xy^2)=-8x^3-2xy+y^3\]\[\frac{d y}{d x}=\frac{-8x^3-2xy+y^3}{x^2-3x^3y^2}\]

Therefore, the value of \[\frac{d y}{d x}\] is \[\frac{-8x^3-2xy+y^3}{x^2-3x^3y^2}\].

Hence, the required answer is \[\frac{-8x^3-2xy+y^3}{x^2-3x^3y^2}\].

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