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Name: How do I find the derivative at the 4 places where x=1 25
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Description: How do I find the derivative at the 4 places where x=1 25

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

Problem 53E

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How do I find the derivative at the 4 places where x=1

Expert Solution

Step 1

Given function is ${(x^{2} + y^{2})}^{2} = \frac{25}{4} x y^{2}$

We have, some rule of differentiation,

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

$\frac{d}{d x} (u v) = v \frac{d}{d x} (u) + u \frac{d}{d x} (v)$

Differentiating the given function with respect to x,

$\begin{array}{rcl} \frac{d}{d x} [{(x^{2} + y^{2})}^{2}] & = & \frac{d}{d x} (\frac{25}{4} x y^{2}) \\ 2 (x^{2} + y^{2}) \cdot \frac{d}{d x} (x^{2} + y^{2}) & = & \frac{25}{4} (x \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} (x)) \\ 2 (x^{2} + y^{2}) \cdot (2 x + 2 y \frac{d y}{d x}) & = & \frac{25}{4} (x \cdot 2 y \frac{d y}{d x} + y^{2}) \\ 4 x (x^{2} + y^{2}) + 4 y (x^{2} + y^{2}) \frac{d y}{d x} & = & \frac{25}{2} x y \frac{d y}{d x} + \frac{25}{4} y^{2} \\ [4 y (x^{2} + y^{2}) - \frac{25}{2} x y] \frac{d y}{d x} & = & \frac{25}{4} y^{2} \\ \frac{d y}{d x} & = & \frac{\frac{25}{4} y^{2}}{4 y (x^{2} + y^{2}) - \frac{25}{2} x y} \end{array}$

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