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How do you find the derivative of #x(t)=4tan^4(2t)#?

Answer 1

Charles Anctil

The derivativeis #x'(t)=32tan^3(2t).sec^2(2t)#

We do a chain derivation #x'(t)=(4tan^4(2t))'=4(4tan^3(2t)).tan(2t)'# using the following

#(x^n)'=nx^(n-1)# #(tanx)'=sec^2x# #(ax)'=a#

So #x'(t)=16tan^3(2t).tan(2t)'#

#=16tan^3(2t).sec^2(2t).(2t)'#

#=16tan^3(2t).sec^2(2t).2#

#=32tan^3(2t).sec^2(2t).#

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Answer 2

Paisley Johnson

To find the derivative of ( x(t) = 4\tan^4(2t) ), you can use the chain rule:

[ \frac{d}{dt} (f(g(t))) = f'(g(t)) \cdot g'(t) ]

Let ( u = 2t ). Then, ( \frac{du}{dt} = 2 ).

Now, differentiate ( \tan^4(u) ) with respect to ( u ) using the chain rule:

[ \frac{d}{du} (\tan^4(u)) = 4\tan^3(u) \cdot \sec^2(u) ]

Since ( u = 2t ), the derivative of ( \tan^4(2t) ) with respect to ( t ) is:

[ \frac{d}{dt} (\tan^4(2t)) = 4\tan^3(2t) \cdot \sec^2(2t) \cdot 2 ]

[ = 8\tan^3(2t) \cdot \sec^2(2t) ]

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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.