How do you differentiate #f(x)=(cosx-x)/(sin3x-x^2)# using the quotient rule?

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Answer 1 · 2017-03-05T10:54:51

#((-sinx-1)(sin3x-x^2)-(3cos3x-2x)(cosx-x))/(sin3x-x^2)^2#

The quotient rule states that

#d/dx f(x)/g(x) = (f'(x)g(x)-g'(x)f(x))/g^2(x)#

Here,

#f(x) = cosx-x#

so

#f'(x) = d/dx[cosx]-d/dx[x] = -sinx-1#

and

#g(x) = sin3x - x^2#

so

#g'(x) = d/dx[sin3x]-d/dx[x^2] = 3cos3x-2x#

We can put these functions into the formula for the quotient rule:

#d/dx f(x)/g(x) = ((-sinx-1)(sin3x-x^2)-(3cos3x-2x)(cosx-x))/(sin3x-x^2)^2#

which you could expand out to give

#= (-sinxsin3x+x^2sinx-sin3x+x^2-3cosxcos3x+3xcos3x+2xcosx-2x^2)/(sin^2(3x)-2x^2sin3x+x^4)#

though this is more complicated and unnecessary.