- #1

darkknight1

- 1

- 0

y = (csc(x) + cot(x) )^-1

Find dy/dx

Find dy/dx

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- MHB
- Thread starter darkknight1
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- #1

darkknight1

- 1

- 0

y = (csc(x) + cot(x) )^-1

Find dy/dx

Find dy/dx

- #2

Monoxdifly

MHB

- 284

- 0

Suppose u = csc(x) + cot x

dy/dx = dy/du × du/dx

dy/dx = dy/du × du/dx

- #3

HOI

- 923

- 2

darkknight, you titled this "using chain rule" and Monoxdifly just told you what that is! Are you able to do this now?y = (csc(x) + cot(x) )^-1

Find dy/dx

Letting u= csc(x)+ cot(x), y= u^-1. What is dy/du? What is du/dx?

chain rule: dy/dx= (dy/du)(du/dx).

If the difficulty is du/dx, it might help you to write \(\displaystyle u= csc(x)+ cot(x)= \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}= \frac{1+ cos(x)}{sin(x)}\) and use the "quotient rule".

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