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How do you differentiate #f(x)=sqrt(1/csc(2/x ) # using the chain rule?

Answer 1

Christopher Allers

#f'(x)=-csc(2/x)/(x^2*sqrt(sin(2/x)))#

Note that #csc(x)=1/sin(x)# so we get by the chain rule

#f'(x)=1/2*(1/csc(2/x))^(-1/2)*csc(x/2)*(-2/x^2)#

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Answer 2

William Abdalla

To differentiate ( f(x) = \sqrt{1/\csc(2/x)} ) using the chain rule, follow these steps:

Let ( u(x) = 1/\csc(2/x) ).
Rewrite ( u(x) ) as ( u(x) = \sin(2/x) ).
Let ( v(x) = \sqrt{u(x)} ).
Apply the chain rule: ( f'(x) = v'(x) = \frac{dv}{du} \cdot \frac{du}{dx} ).
Find ( \frac{dv}{du} ) and ( \frac{du}{dx} ).
( \frac{dv}{du} = \frac{1}{2\sqrt{u(x)}} ).
( \frac{du}{dx} = \frac{d}{dx}[\sin(2/x)] = \cos(2/x) \cdot \frac{d}{dx}(2/x) ).
Use the chain rule again for ( \frac{d}{dx}(2/x) ): ( \frac{d}{dx}(2/x) = -\frac{2}{x^2} ).
Combine the results: ( f'(x) = \frac{1}{2\sqrt{1/\csc(2/x)}} \cdot \cos(2/x) \cdot \left(-\frac{2}{x^2}\right) ).
Simplify the expression for ( f'(x) ): ( f'(x) = -\frac{\cos(2/x)}{x^2\sqrt{\csc(2/x)}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.