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How do you differentiate #f(x)= ( -2x^2+ 3x ) / ( - e^x + 2 ) # using the quotient rule?

Answer 1

Cameron Alexander

#(df)/(dx)=(-e^x(2x^2-7x+3)-8x+6)/(-e^x+2)^2#

Quotient rule states that if #f(x)=g(x)/(h(x))#, then

#f'(x)=(h(x)*g'(x)-h'(x)*g(x))/(h(x)^2)#

Hence for #f(x)=(-2x^2+3x)/(-e^x+2)#

#(df)/(dx)=((-e^x+2)(-4x+3)-(-e^x)(-2x^2+3x))/(-e^x+2)^2#

= #(4xe^x-3e^x-8x+6-2x^2e^x+3xe^x)/(-e^x+2)^2# or

= #(-e^x(2x^2-7x+3)-8x+6)/(-e^x+2)^2#

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Answer 2

Charles Arocha

To differentiate ( f(x) = \frac{-2x^2 + 3x}{-e^x + 2} ) using the quotient rule, follow these steps:

Identify ( u(x) ) and ( v(x) ): ( u(x) = -2x^2 + 3x ) ( v(x) = -e^x + 2 )
Find ( u'(x) ) and ( v'(x) ): ( u'(x) = -4x + 3 ) ( v'(x) = -e^x )
Apply the quotient rule: ( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} )
Substitute the values: ( f'(x) = \frac{(-e^x + 2)(-4x + 3) - (-2x^2 + 3x)(-e^x)}{(-e^x + 2)^2} )
Simplify the expression as needed.

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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.