Consider the following. 8x² + y² = 5x (a) Find y' by implicit differentiation. - y' = 5-32x 71,6 3 (b) Solve the equation explicitly for y and differentiate to get y' in terms of x. 5-321³ 75x8) y' = 6 X (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a).
Consider the following. 8x² + y² = 5x (a) Find y' by implicit differentiation. - y' = 5-32x 71,6 3 (b) Solve the equation explicitly for y and differentiate to get y' in terms of x. 5-321³ 75x8) y' = 6 X (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a).
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...
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![**Consider the following:**
\[ 8x^4 + y^7 = 5x \]
**(a)** Find \( y' \) by implicit differentiation.
\[ y' = \frac{5 - 32x^3}{7y^6} \] ✔️
**(b)** Solve the equation explicitly for \( y \) and differentiate to get \( y' \) in terms of \( x \).
\[ y' = \frac{5 - 32x^3}{7(5x - 8x^4)^{\frac{6}{7}}} \] ❌
**(c)** Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \( y \) into your solution for part (a).
\[ y' = \_\_\_\_\_\_\_ \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8da5c625-81d8-4db3-9f61-2b28dd7af5ea%2Fab8d3e19-c565-45d8-b773-ed7313d775d7%2Fv1s99sm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Consider the following:**
\[ 8x^4 + y^7 = 5x \]
**(a)** Find \( y' \) by implicit differentiation.
\[ y' = \frac{5 - 32x^3}{7y^6} \] ✔️
**(b)** Solve the equation explicitly for \( y \) and differentiate to get \( y' \) in terms of \( x \).
\[ y' = \frac{5 - 32x^3}{7(5x - 8x^4)^{\frac{6}{7}}} \] ❌
**(c)** Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \( y \) into your solution for part (a).
\[ y' = \_\_\_\_\_\_\_ \]
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