How do you differentiate #f(x)=1/(cot(x)) # using the chain rule?

Question

How do you differentiate #f(x)=1/(cot(x))   # using the chain rule?

Emma Aldridge · Accepted Answer

#(cosec^2 x)/(cot^2 x)# differentiate using the #color(blue)" chain rule " # #d/dx [ f(g(x)) ] = f'(g(x) . g'(x) # and the standard derivative D(cotx) = #- cosec^2 x # #"---------------------------------------------------------------------------"# rewrite f(x) as f(x) # = (cotx)^-1 # f(g(x)) =#(cotx)^-1 rArr f'(g(x)) = -1(cotx)^-2 # and g(x) = # cotx rArr g'(x) = -cosec^2 x # #rArr f'(x) = -(cotx)^-2 . - cosec^2 x = (cosec^2 x)/(cot^2 x) #

Emilia Aguilar · Answer

undefined To differentiate $ f(x) = \frac{1}{\cot(x)} $ using the chain rule, follow these steps:

1. Recognize that $ f(x) $ can be rewritten as $ f(x) = \cfrac{1}{\tan(x)} $, since $ \cot(x) = \frac{1}{\tan(x)} $.
2. Apply the chain rule. Let $ u(x) = \tan(x) $. Then $ f(x) = \frac{1}{u(x)} $.
3. Differentiate $ u(x) = \tan(x) $ with respect to $ x $ to find $ u'(x) $.
4. Use the chain rule formula: $ \frac{d}{dx} \left( \frac{1}{u(x)} \right) = -\frac{1}{u(x)^2} \cdot u'(x) $.
5. Substitute $ u(x) = \tan(x) $ and $ u'(x) = \sec^2(x) $.
6. The result is $ f'(x) = -\frac{\sec^2(x)}{\tan^2(x)} $.

So, the derivative of $ f(x) = \frac{1}{\cot(x)} $ using the chain rule is $ f'(x) = -\frac{\sec^2(x)}{\tan^2(x)} $.