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

1 Answer
mason m
Mar 19, 2016

#f'(x)=(2/x^3+1)csc(1/x^2-x)cot(1/x^2-x)#

Explanation:

The derivative of #csc(x)# is:

#d/dxcsc(x)=-csc(x)cot(x)#

The chain rule states that when differentiating a function inside another function, differentiate the outside function while leaving the inside function intact, and then multiply that by the derivative of the inside function.

When applied to #csc(u)#, this can be shown as:

#d/dxcsc(u)=-csc(u)cot(u)*u'#

Here, #u=1/x^2-x#, so

#u=x^-2-x" "=>" "u'=-2x^-3-1#

#u'=-2/x^3-1#

Thus,

#d/dxcsc(1/x^2-x)=-csc(1/x^2-x)cot(1/x^2-x)*(-2/x^3-1)#

Distributing the negative,

#=(2/x^3+1)csc(1/x^2-x)cot(1/x^2-x)#

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