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How do you differentiate #f(x)=e^x/x-3x# using the quotient rule?

Answer 1

Elijah Abram

You'll be using two different differentiating rules for this function.

Quotient and multiplication by constant:

#f'(x)=[d/dx(e^x)*x-d/dx(x)*e^x]/x^2-d/dx(3x)#

#=[xe^x-e^x]/x^2-3#

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Answer 2

Evelyn Agard

To differentiate the function ( f(x) = \frac{e^x}{x - 3x} ) using the quotient rule, you apply the formula:

[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} ]

where ( u = e^x ) and ( v = x - 3x ).

Thus, ( u' = e^x ) and ( v' = 1 - 3 = -2 ).

Substituting into the formula:

[ f'(x) = \frac{(e^x)(x - 3x) - (e^x)(1 - 3)}{(x - 3x)^2} ] [ = \frac{(e^x)(x - 3x) - (e^x)(1 - 3)}{(1 - 3x)^2} ] [ = \frac{(e^x)(x - 3x) - (e^x)(-2)}{(1 - 3x)^2} ] [ = \frac{xe^x - 3xe^x + 2e^x}{(1 - 3x)^2} ] [ = \frac{(xe^x - 3xe^x) + 2e^x}{(1 - 3x)^2} ] [ = \frac{-2xe^x + 2e^x}{(1 - 3x)^2} ] [ = \frac{2e^x(-x + 1)}{(1 - 3x)^2} ]

So, ( f'(x) = \frac{2e^x(-x + 1)}{(1 - 3x)^2} ).

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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.