How do you differentiate #arcsin(csc(4/x)) )# using the chain rule?

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Answer 1 · 2016-02-03T00:38:55

#f'(x)=(4csc(4/x)cot(4/x))/(x^2sqrt(1-csc^2(4/x))#

To differentiate an #arcsin# function, use the chain rule:

#d/dx[arcsin(u)]=1/sqrt(1-u^2)*u'#

Here, #u=csc(4/x)#, so we see that

#f'(x)=1/(sqrt(1-csc^2(4/x)))*d/dx[csc(4/x)]#

Now, to differentiate the #csc# function, use the chain rule again:

#d/dx[csc(u)]=-csc(u)cot(u)*u'#

Since this time #u=4/x#,

#f'(x)=1/sqrt(1-csc^2(4/x))*-csc(4/x)cot(4/x)*d/dx[4/x]#

Use the product rule to see that #d/dx[4/x]=d/dx[4x^-1]=-4x^-2=-4/x^2#, giving a final simplified derivative of

#f'(x)=(4csc(4/x)cot(4/x))/(x^2sqrt(1-csc^2(4/x))#