Differentiate the following equation

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SUMMARY

The discussion focuses on differentiating the function s = Tan²(e^(4t)). The correct derivative is established as s' = 8e^(4t) * Tan(e^(4t)) * Sec²(e^(4t)). The user initially misinterpreted the problem, thinking they needed the derivative itself instead of evaluating it at a specific point, s'(0). The evaluation at t=0 yields v(0) = 8 * Tan(1) / Cos²(1), which is confirmed as correct by other participants.

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Homework Statement



Differentiate

s = Tan ^{2}(e^{4t})



Homework Equations



[e^{u}]' = u' e^{u}

[Tan(θ)]' = Sec^{2}(θ)

Sec(θ) = \frac{1}{Cos(θ)}



The Attempt at a Solution



s' = 2[Tan(e^{4t})] * Sec^{2}(e^{4t}) * 4e^{4t}

s' = 8e^{4t} * Tan(e^{4t}) * Sec^{2}(e^{4t})

s' = 8e^{4t} * \frac{Tan(e^(4t))}{Cos^2(e^(4t))}



Mathway says that s' = 0... Can someone please let me know where I went wrong? Thanks
 
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You're derivative is right. Are you sure they didn't ask you to evaluate the derivative at some point? I'm guessing that they want you to evaluate it at 0.
 
Yes, you're right, I have to find s'(0) aka v(0)

I end up getting this:

v(0) = 8e^{0} * \frac{Tan(e^0)}{Cos^2(e^0)}

v(0) = 8 * \frac{Tan(1)}{Cos^2(1)}

Is that good?
 
Blablablabla said:
Yes, you're right, I have to find s'(0) aka v(0)

I end up getting this:

v(0) = 8e^{0} * \frac{Tan(e^0)}{Cos^2(e^0)}

v(0) = 8 * \frac{Tan(1)}{Cos^2(1)}

Is that good?

Yes, it is.
 

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