How do you differentiate #f(x)=8e^(x^2)/(e^x+1)# using the chain rule?
The only trick here is that
Final derivative is:
or
Note: if you want to study the sign, you are gonna have a bad time. Just look at the graph:
graph{8(e^(x^2))/(e^x+1) [-50.25, 53.75, -2.3, 49.76]}
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To differentiate ( f(x) = \frac{8e^{x^2}}{e^x + 1} ) using the chain rule, follow these steps:
- Identify the outer function ( u ) and the inner function ( v ).
- Compute ( u' ) and ( v' ).
- Apply the chain rule: ( \frac{d}{dx}(u(v(x))) = u'(v(x)) \cdot v'(x) ).
- Substitute ( u' ), ( v' ), and ( v(x) ) into the chain rule formula.
- Simplify the expression.
Here's the step-by-step process:
- Let ( u(x) = \frac{8e^x}{e^x + 1} ) and ( v(x) = x^2 ).
- Compute ( u'(x) = \frac{8(e^x)(e^x + 1) - 8e^x(e^x)}{(e^x + 1)^2} ) and ( v'(x) = 2x ).
- Apply the chain rule: ( \frac{d}{dx}(u(v(x))) = u'(v(x)) \cdot v'(x) ).
- Substitute ( u' ), ( v' ), and ( v(x) ): ( \frac{d}{dx}\left(\frac{8e^{x^2}}{e^x + 1}\right) = \frac{8(e^{x^2})(2x)}{e^{x^2} + 1} ).
- Simplify the expression: ( \frac{16xe^{x^2}}{e^{x^2} + 1} ).
Therefore, ( \frac{d}{dx}\left(\frac{8e^{x^2}}{e^x + 1}\right) = \frac{16xe^{x^2}}{e^{x^2} + 1} ).
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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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