How do you differentiate #f(x)= e^x/(x-7 )# using the quotient rule?

Answer 1

#f'(x)=((x-6)e^x)/(x-7)^2#

Quotient rule states that

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

Application to the given examples yields

#f'(x)=((x-7)e^x+e^x)/(x-7)^2#
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Answer 2

To differentiate ( f(x) = \frac{e^x}{x - 7} ) using the quotient rule:

  1. Let ( u = e^x ) and ( v = x - 7 ).
  2. Find ( u' ) and ( v' ).
    • ( u' = e^x ) (since the derivative of ( e^x ) is ( e^x ))
    • ( v' = 1 ) (since the derivative of ( x - 7 ) is ( 1 ))
  3. Apply the quotient rule: [ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ]
  4. Substitute ( u' ), ( v' ), ( u ), and ( v ) into the formula: [ \frac{d}{dx}\left(\frac{e^x}{x - 7}\right) = \frac{(e^x)(x - 7) - (e^x)(1)}{(x - 7)^2} ]
  5. Simplify the expression: [ \frac{d}{dx}\left(\frac{e^x}{x - 7}\right) = \frac{e^x(x - 7) - e^x}{(x - 7)^2} ]
  6. Further simplification may be possible, but this is the result of differentiating ( f(x) ) using the quotient rule.
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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.

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.

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.

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.

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