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

Answer 1

Ezra Adams

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

As per the quotient rule,

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

Here,

#f(x) = cosx-x#

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

and

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

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

These functions can be added to the quotient rule formula as follows:

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

which you could further develop to provide

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

Nevertheless, this is needless and more complicated.

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Answer 2

William Abdalla

To differentiate ( f(x) = \frac{\cos(x) - x}{\sin(3x) - x^2} ) using the quotient rule:

Apply the quotient rule: [ f'(x) = \frac{( \sin(3x) - x^2 ) \cdot (-\sin(x)) - (\cos(x) - x) \cdot (3\cos(3x) - 2x)}{(\sin(3x) - x^2)^2} ]
Simplify the expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.