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

Answer 1

#\frac{(e^{x^2}*2x)(e^x-x^2)-e^{x^2}(e^x-2x)}{(e^x-x^2)^2}#

Rule of the quotient:

#d/dx f(x)/g(x) = \frac{f'g-fg'}{g^2}#

Let's calculate every function that we require:

Now, enter these numbers into the formula:

#\frac{f'g-fg'}{g^2} = \frac{(e^{x^2}*2x)(e^x-x^2)-e^{x^2}(e^x-2x)}{(e^x-x^2)^2}#

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Answer 2

To differentiate ( f(x) = \frac{e^{x^2}}{e^x - x^2} ) using the quotient rule, first identify ( u(x) = e^{x^2} ) and ( v(x) = e^x - x^2 ). Then, the derivative is calculated as follows:

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

Where ( u' ) and ( v' ) denote the derivatives of ( u(x) ) and ( v(x) ) respectively.

( u(x) = e^{x^2} ) implies ( u'(x) = 2x \cdot e^{x^2} ) (by chain rule), and ( v(x) = e^x - x^2 ) implies ( v'(x) = e^x - 2x ).

Plugging these into the quotient rule formula:

[ f'(x) = \frac{(2x \cdot e^{x^2})(e^x - x^2) - (e^{x^2})(e^x - 2x)}{(e^x - x^2)^2} ]

This simplifies to:

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

Combining like terms, the final result is:

[ f'(x) = \frac{4xe^{x^2} - e^{2x^2} - 2x^3e^{x^2}}{(e^x - x^2)^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.

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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