"Find the derivatives of the function y defined..." was answered by our community

Teddy Dillard

Answered question

2021-12-17

Find the derivatives of the function y defined implicity by each of the following equation

x^{4} + y^{4} - a^{2} x y = 0

Thomas Lynn

Beginner2021-12-18Added 28 answers

Step 1
Given

x^{4} + y^{4} - a^{2} x y = 0

Step 2
differentiating wrt "x"

\frac{d}{d x} (x^{4}) + \frac{d}{d x} (y^{4}) - a^{2} \frac{d}{d x} (x y) = 0

4 x^{3} + 4 y^{3} \frac{d y}{d x} - a^{2} (x \frac{d y}{d x} + y \cdot 1) = 0

4 x^{3} + 4 y^{3} \frac{d y}{d x} - a^{2} x \frac{d y}{d x} - a^{2} y = 0

(4 y^{3} - a^{2} x) \frac{d y}{d x} = a^{2} y - 4 x^{3}

∴ \frac{d y}{d x} = \frac{a^{2} y - 4 x^{2}}{4 y^{3} - a^{2} x}

John Koga

Beginner2021-12-19Added 33 answers

Step 1
In implicit differentiation all the differentiation rules are applicable.
Step 2

x^{4} + y^{4} - a^{2} x y = 0

differentiate both sides wrt x:

\frac{d}{d x} (x^{4} + y^{4} - a^{2} x y) = \frac{d}{d x} (0)

\frac{d}{d x} (x^{4}) + \frac{d}{d x} (y^{4}) - a^{2} \frac{d}{d x} (x y) = 0

apply chain rule and product rule of differentiation:

4 x^{3} + 4 y^{3} \frac{d y}{d x} - a^{2} (x \frac{d y}{d x} + y \frac{d x}{d x}) = 0

4 x^{3} + 4 y^{3} y^{'} - a^{2} (x y^{'} + y) = 0

4 x^{3} + 4 y^{3} y^{'} - a^{2} x y^{'} - a^{2} y = 0

y^{'} (4 y^{3} - a^{2} x) = a^{2} y - 4 x^{3}

y^{'} = \frac{a^{2} y - 4 x^{3}}{4 y^{3} - a^{2} x}

Final answer:

y^{'} = \frac{a^{2} y - 4 x^{3}}{4 y^{3} - a^{2} x}

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