Do you use the chain rule for the derivative of sec(2x)?

Question

If you don't, wouldnt it be sec2xtan2x?

But if you do, and its in a problem where u already used the chain rule, you do the chain rule again?

ex.) (sec2x)/(1+tan2x)

Ok I'm assuming you used the chain rule..
thanks

Answer 1

You have to use the chain rule because 2x is still a function of x.
the derivative of sec(2x)
= sec(2x)tan(2x)(2)

Answer 2

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RE:
Do you use the chain rule for the derivative of sec(2x)?
If you don't, wouldnt it be sec2xtan2x?

But if you do, and its in a problem where u already used the chain rule, you do the chain rule again?

ex.) (sec2x)/(1+tan2x)

Answer 3

You do for the 2x part.
The derivative is:

sec2xtan2x (2) = 2 sec2x tan2x

Answer 4

Derivative Of Sec2x

Answer 5

I think d/dx = 2 sec 2x tan 2x

but it's been a long time!

Answer 6

It appears to be:
2Sec[2x]Tan[2x]
according to this source:
http://integrals.wolfram.com/index.jsp
The Wolfram Integrator

(but I don't know why)

Answer 7

chain rule derivative sec 2x: https://tinyurl.im/e/do-you-use-the-chain-rule-for-the-derivative-of-sec-2x

Answer 8

y = sec 2x = 1/cos2x

Derivative y = u/v

y' = [(cos2x)(0) - (1)(-2sin2x)]/ [cos^2 (2x)]
y' = 2sin2x / [cos^2 (2x)]
y' = 2[sin2x / cos2x] [1/cos2x]
y' = 2 tan2x sec2x

Answer 9

This is THE question regarding the chain rule.
First, keep in mind that: dx/dx = 1
now let v=g(u) and u=f(x)
dv/dv = 1
dv/du = dv/du * du/du = dv/du * 1 = v'(u)
dv/dx = dv/du * du/dx * dx/dx = dv/du * du/dx * 1 = v'(u) * u'(x)

In your problem, your v = g(u) is
v=sec(u) u=2x
so do dv/du -or v'(u), or d/du sec(u) - times du/dx -or u'(x) or d/dx (2x) -

Answer 10

I make it a practice to avoid the chain rule at all costs. Now the platypus rule, that is the one which should be applied for your problem.