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

Question

William Abdalla · Accepted Answer

#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#

Emma Aldridge · Answer

undefined 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} $

Scarlett Allison · Answer

undefined 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} $.