How do you differentiate #f(x)= ( x sinx )/ ( x - 6)# using the quotient rule?

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Answer 1 · 2016-01-22T23:21:45

#f'(x)=((x^2-6x)cosx-6sinx)/(x-6)^2#

The quotient rule states that

#f'(x)=((x-6)color(red)(d/dx[xsinx])-xsinxcolor(blue)(d/dx[x-6]))/(x-6)^2#

Now, find both the internal derivatives.

The first requires product rule:

#d/dx[xsinx]=sinxd/dx[x]+xd/dx[sinx]=color(red)(sinx+xcosx#

The second is very simple.

#d/dx[x-6]=color(blue)1#

Plug these back in.

#f'(x)=((x-6)(sinx+xcosx)-xsinx)/(x-6)^2#

Simplify.

#f'(x)=(xsinx+x^2cosx-6sinx-6xcosx-xsinx)/(x-6)^2#

#f'(x)=((x^2-6x)cosx-6sinx)/(x-6)^2#