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

Answer 1

Isabella Ackerman

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

As per the quotient rule,

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

Here,

#f(x) = sinx" "" "" "=>" "" " f'(x) = cosx#

#g(x) = x-e^x" "" "color(white)"-" => " "" "g'(x) = 1 - e^x#

Consequently, if we apply the quotient rule to differentiate, we obtain:

#d/dx(sinx/(x-e^x)) = ((x-e^x)(cosx) - (sinx)(1-e^x))/(x-e^x)^2#

#= (xcosx - e^xcosx - sinx + e^xsinx)/(x-e^x)^2#

Although this can be simplified in other ways, each version is equally accurate.

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Answer 2

Aurora Alvarado

To differentiate f(x) = (sinx)/(x - e^x) using the quotient rule, first identify f'(x) = [g(x)*f'(x) - f(x)*g'(x)] / [g(x)]^2, where f(x) = sinx, g(x) = x - e^x, f'(x) = cosx, and g'(x) = 1 - e^x. Then substitute these values into the quotient rule formula and simplify to find the derivative of f(x).

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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.