Using Chain rule to find derivatives....

In summary, the conversation is discussing finding the derivative of y = (csc(x) + cot(x) )^-1 using the chain rule. The person explains that letting u = csc(x) + cot(x), the derivative can be found by the formula dy/dx = (dy/du)(du/dx). They suggest simplifying u to (1+cos(x))/sin(x) and using the quotient rule to find du/dx.
  • #1
y = (csc(x) + cot(x) )^-1
Find dy/dx
 
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  • #2
Suppose u = csc(x) + cot x

dy/dx = dy/du × du/dx
 
  • #3
darkknight said:
y = (csc(x) + cot(x) )^-1
Find dy/dx
darkknight, you titled this "using chain rule" and Monoxdifly just told you what that is! Are you able to do this now?

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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