How do you find the derivative of #y=ln(1+e^(2x))#?
Cameron AlexanderBy signing up, you agree to our Terms of Service and Privacy Policy
Evelyn AgardTo find the derivative of ( y = \ln(1 + e^{2x}) ), you can use the chain rule. The derivative is:
[ \frac{dy}{dx} = \frac{1}{1 + e^{2x}} \cdot \frac{d}{dx}(1 + e^{2x}) ]
Applying the chain rule to ( 1 + e^{2x} ):
[ \frac{d}{dx}(1 + e^{2x}) = 0 + e^{2x} \cdot \frac{d}{dx}(2x) ]
[ = e^{2x} \cdot 2 ]
[ = 2e^{2x} ]
So, putting it back into the original expression:
[ \frac{dy}{dx} = \frac{1}{1 + e^{2x}} \cdot 2e^{2x} ]
[ = \frac{2e^{2x}}{1 + e^{2x}} ]
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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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