How do you differentiate # g(x) =(1+cosx)/(1-cosx)

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Question

How do you differentiate # g(x) =(1+cosx)/(1-cosx)
 
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Cameron Alexander · Accepted Answer

Use the quotient rule :

#(d(f(x)/(h(x))))/dx = (f'(x)h(x)-f(x)h'(x))/(h(x))^2# Let #f(x) = 1 +cos(x)#, then #f'(x) = -sin(x)# Let #h(x) = 1-cos(x)#, then #h'(x) = sin(x)# Substituting into the quotient rule: #(d((1 +cos(x))/(1-cos(x))))/dx = ((-sin(x))(1-cos(x))-(1 +cos(x))(sin(x)))/(1-cos(x))^2#

Scarlett Allison · Answer

By simplification

#d/dxg(x)=d/dxcot^2(x/2)# #2cot(x/2)*cosec^2(x/2)*1/2# #cot(x/2)*cosec^2(x/2)#

Isabella Ackerman · Answer

#g'(x)=-(2sinx)/(1-cosx)^2# differentiate using the #color(blue)"quotient rule"# #"Given " f(x)=(g(x))/(h(x))" then"# #f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"# #"here " g(x)=1+cosxrArrg'(x)=-sinx# #h(x)=1-cosxrArrh'(x)=sinx# #rArrf'(x)=((1-cosx)(-sinx)-(1+cosx)(sinx))/(1-cosx)^2# #color(white)(rArrf'(x))=(-sinx+sinxcosx-sinx-sinxcosx)/(1-cosx)^2# #color(white)(rArrf'(x))=-(2sinx)/(1-cosx)^2#

Ezekiel Alexander · Answer

undefined To differentiate the function $ g(x) = \frac{1 + \cos x}{1 - \cos x} $, you can use the quotient rule. The quotient rule states that for a function $ f(x) = \frac{u(x)}{v(x)} $, its derivative is given by $ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $. Applying this rule to $ g(x) $, where $ u(x) = 1 + \cos x $ and $ v(x) = 1 - \cos x $, we have:

$$
g'(x) = \frac{(1 + \cos x)(-\sin x) - (1 - \cos x)(-\sin x)}{(1 - \cos x)^2}
$$

Simplify the expression:

$$
g'(x) = \frac{-\sin x - \cos x \sin x + \sin x + \cos x \sin x}{(1 - \cos x)^2}
$$

$$
g'(x) = \frac{2\sin x}{(1 - \cos x)^2}
$$

So, the derivative of $ g(x) $ with respect to $ x $ is $ \frac{2\sin x}{(1 - \cos x)^2} $.

How do you differentiate # g(x) =(1+cosx)/(1-cosx) #?