How do you derive #y = (x-1)/sin(x)# using the quotient rule?

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Answer 1 · 2015-11-06T22:24:51

The derivative is #\frac{sin(x) - (x-1)cos(x)}{sin^2(x)}#

The rule states that

#(f(x)/g(x))' = \frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}#.

So, as long as we know #f(x),g(x),f'(x),g'(x)# and #g^2(x)#, we're ready to write the derivative. Let's compute this quantities:

#f(x)# is of course #x-1#, and #g(x)# is #sin(x)#.

This means that #f'(x)=1#, and #g'(x)=cos(x)#.

Finally, #g^2(x)=sin^2(x)#

Now we have all the "ingredients", and we can put them in the "recipe":

#\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} -> \frac{1*sin(x) - (x-1)cos(x)}{sin^2(x)}#