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

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Answer 1 · 2015-11-14T00:27:16

#d/dx1/(sin(x) + cos(x))^2 = (2(sin(x)-cos(x)))/(sin(x)+cos(x))^3#

In this problem, we will use the following:
The chain rule:
#d/dxf(g(x)) = f'(g(x))g'(x)#

The power rule:
#d/dxx^n = nx^(n-1)#

and the following derivatives:
#d/dx sin(x) = cos(x)#
#d/dx cos(x) = -sin(x)#

To avoid needing the quotient rule, we first write
#d/dx1/(sin(x) + cos(x))^2 = d/dx(sin(x)+cos(x))^-2#

Next, by the chain rule and power rule, we have
#d/dx(sin(x)+cos(x))^-2 = -2(sin(x)+cos(x))^-3(d/dx(sin(x)+cos(x)))#

#=>d/dx(sin(x)+cos(x))^-2 =#
#-2(sin(x)+cos(x))^-3(cos(x)-sin(x))#

so, bringing the #-1# into the numerator, our final result is

#d/dx1/(sin(x) + cos(x))^2 = (2(sin(x)-cos(x)))/(sin(x)+cos(x))^3#