Question

cot^-1
(4-x-2x^2/3x+2) find Dy/dx
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Answer 1

Answer:

Sorry guy

I don't know the answer

Answer 2

Explanation:

Given Cot^-1(4-x-2x^2/3x+2) find Dy/dx

•     Let y = Cot^-1(4-x-2x^2/3x+2)
•            Since  
•                      cot^-1 x = tan^-1 1/x
•         So we can write this as  
•                      y = tan^-1 (3x + 2) / (4 – x – 2x^2)
•                   So we can write this as  
•                      y = tan^-1 2x + x + 3 – 1 / 1 + 3 – x – 2x^2
•                     y = 2x + x + 3 – 1 / 1 – (2x^2 + x – 3)
•                      y = 2x + x + 3 – 1 / 1 – (2x^2 + 3x – 2x – 3)
•                     y = 2x + x + 3 – 1 / 1 – ( 2x + 3) (x – 1)
•                        So we have
•                       y = (2x + 3) + (x – 1) / 1 – (2x + 3) (x – 1)
•                   So we get
•                        y = tan^-1 (2x + 3) + tan^-1 (x – 1)
•                     Differentiating with respect to x we get
•                         dy / dx = d/dx (tan^-1 (2x + 3) + tan^-1 (x – 1)
•                         Now d/dx (tan^-1 2x + 3)  
•                            = 1/1 + (2x + 3)^2 . d/dx (2x + 3)
•                             = 1/1 + (2x + 3)^2 (2 x 1 + 0)
•                                = So we get 2 / 1 + (2x + 3)^2
•                         Now d/dx (tan^-1 (x – 1))
•                                = 1 / 1 + (x – 1)^2 d/dx (x – 1)
•                                 = 1 / 1 + (x – 1)^2 (1 – 0)
•                                   = so we get  1 / 1 + (x – 1)^2
• So dy / dx = 2 / 1 + (2x + 3)^2 + 1 / 1 + (x – 1)^2

Reference link will be

https://brainly.in/question/15531711