`f(x) = 1/(1 + sec(x))^2` Find the derivative of the function.

Expert Answers
hkj1385 eNotes educator| Certified Educator

Note:- 1) If y = x^n ; then dy/dx = n*{x^(n-1)}

2) If y = secx ; then dy/dx = secx*tanx

Now, 

f(x) = y = 1/[{1+secx}^2] = {1+secx}^(-2)

thus, dy/dx = f'(x) = y' = (-2)*[{1+secx}^(-3)]*(secx*tanx)

or, dy/dx = f'(x) = y' = (-2*secx*tanx)/[{1+secx}^(3)]

balajia | Student

The given function is `f(x)=(1+secx)^(-2)`

`f'(x)=(-2)(1+secx)^(-2-1)(secx.tanx)`

`f'(x)=(-2secx.tanx)/(1+secx)^3`

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