How do you differentiate #f(x)=csc(sqrt(x^2-5x)) # using the chain rule?

Answer 1

#f'(x)=-(2x-5)/(2sqrt (x^2-5x))csc (sqrt (x^2-5x))cot (sqrt (x^2-5x))#

#f (x)=csc(sqrt (x^2-5x))# Differentiating both sides with respect to #x# we get #f'(x)=-csc (sqrt (x^2-5x)).cot (sqrt (x^2-5x))×d/dx (sqrt (x^2-5x))# #=>f'(x)=-csc (sqrt (x^2-5x)).cot (sqrt (x^2-5x))×1/(2sqrt (x^2-5x))×(2x-5)# #:.f'(x)=-(2x-5)/(2sqrt (x^2-5x))csc (sqrt (x^2-5x))cot (sqrt (x^2-5x))#

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Answer 2

Scarlett Allison

To differentiate ( f(x) = \csc(\sqrt{x^2 - 5x}) ) using the chain rule, we proceed as follows:

[ f'(x) = -\csc(\sqrt{x^2 - 5x}) \cot(\sqrt{x^2 - 5x}) \frac{d}{dx}(\sqrt{x^2 - 5x}) ]

[ = -\csc(\sqrt{x^2 - 5x}) \cot(\sqrt{x^2 - 5x}) \frac{1}{2\sqrt{x^2 - 5x}} \frac{d}{dx}(x^2 - 5x) ]

[ = -\csc(\sqrt{x^2 - 5x}) \cot(\sqrt{x^2 - 5x}) \frac{1}{2\sqrt{x^2 - 5x}} (2x - 5) ]

[ = -\frac{x - 5}{\sqrt{x^2 - 5x}} \cot(\sqrt{x^2 - 5x}) \csc(\sqrt{x^2 - 5x}) ]

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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.