# What is the derivative of # f(x)=1/(sinx+cosx)^2#?

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To find the derivative of ( f(x) = \frac{1}{{(\sin x + \cos x)}^2} ), you can use the chain rule and the quotient rule.

( f'(x) = -2 \frac{\sin x + \cos x}{(\sin x + \cos x)^3} (\cos x - \sin x) )

Simplified:

( f'(x) = -\frac{2(\cos x - \sin x)}{(\sin x + \cos x)^3} )

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The derivative of ( f(x) = \frac{1}{(\sin(x) + \cos(x))^2} ) can be found using the chain rule and the derivative of the inner function (\sin(x) + \cos(x)).

Let ( u = \sin(x) + \cos(x) ).

Then, ( \frac{du}{dx} = \cos(x) - \sin(x) ).

Now, applying the chain rule:

( \frac{d}{dx}\left(\frac{1}{u^2}\right) = -\frac{1}{u^2} \cdot \frac{du}{dx} )

Substituting back ( u = \sin(x) + \cos(x) ) and ( \frac{du}{dx} = \cos(x) - \sin(x) ):

( = -\frac{1}{(\sin(x) + \cos(x))^2} \cdot (\cos(x) - \sin(x)) )

So, the derivative of ( f(x) = \frac{1}{(\sin(x) + \cos(x))^2} ) is ( -\frac{\cos(x) - \sin(x)}{(\sin(x) + \cos(x))^2} ).

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