Differentiate the following equation

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Blablablabla
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Homework Statement



Differentiate

s = Tan [itex]^{2}[/itex](e[itex]^{4t}[/itex])



Homework Equations



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

[Tan(θ)]' = Sec[itex]^{2}[/itex](θ)

Sec(θ) = [itex]\frac{1}{Cos(θ)}[/itex]



The Attempt at a Solution



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

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

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



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[itex]^{0}[/itex] * [itex]\frac{Tan(e^0)}{Cos^2(e^0)}[/itex]

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

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[itex]^{0}[/itex] * [itex]\frac{Tan(e^0)}{Cos^2(e^0)}[/itex]

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

Is that good?

Yes, it is.