Find dy/dx by implicit differentiation. x6 + y² = 23 Step 1 With implicit differentiation, we differentiate all the terms in our equation with respect to x. Keep in mind we are implying that y is a function of x. Think of it as y(x). In order to differentiate it, we will need to use the Chain Rule. To start, differentiate both sides of the equation with respect to x. x6 + y² = 23 [x6] + dx + + [3] = [23] [³] = 0 Step 2 Use the Chain Rule to find the derivative of y3. If it helps, think of y as y(x). [y³] = dy dx 0 X

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
Find dy/dx by implicit differentiation.
x6 + y³ = 23
Step 1
With implicit differentiation, we differentiate all the terms in our equation with respect to x. Keep in mind we are implying that y is a function of x. Think of it as y(x). In order to
differentiate it, we will need to use the Chain Rule.
To start, differentiate both sides of the equation with respect to x.
x6 + y³ = 23
d_[y³]
dx
6 x
-[x6] +
dx
5
+
d_[y³]
dx
=
=
[23]
dx
Step 2
Use the Chain Rule to find the derivative of y³. If it helps, think of y as y(x).
dy
[³] =
dx
Transcribed Image Text:Find dy/dx by implicit differentiation. x6 + y³ = 23 Step 1 With implicit differentiation, we differentiate all the terms in our equation with respect to x. Keep in mind we are implying that y is a function of x. Think of it as y(x). In order to differentiate it, we will need to use the Chain Rule. To start, differentiate both sides of the equation with respect to x. x6 + y³ = 23 d_[y³] dx 6 x -[x6] + dx 5 + d_[y³] dx = = [23] dx Step 2 Use the Chain Rule to find the derivative of y³. If it helps, think of y as y(x). dy [³] = dx
Expert Solution
Step 1: Formulas of derivative

Given, x to the power of 6 plus y cubed equals 23

formulas of derivatives,

(i) fraction numerator d space over denominator d x end fraction left parenthesis x to the power of n right parenthesis equals n x to the power of n minus 1 end exponent

(ii) fraction numerator d space over denominator d x end fraction left parenthesis c o n s tan t right parenthesis equals 0

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