Question

Find g '(x) for g(x) = sec(2x).

a) g '(x) = 2sec^2(2x)
b) g '(x) = 2sec(2x)
c) g '(x) = 2sec(2x)tan(2x)
d) g '(x) = 2tan^2(2x)

I think i got: C? i just want someone to check my answer :)

Answer 1

@ganeshie8

Answer 2

i think its c

Answer 3

This involves using the chain rule: \(\frac{d}{dx}(\sec(x) \cdot \frac{d}{dx}(x)\)

Answer 4

2sec(2x)tan(2x)

Answer 5

c is the answer

Answer 6

that's what i got too @laetitiacarmel :) and @Jhannybean ok, do u think its correct as C? :)

Answer 7

yes

Answer 8

Do you know your derivatives @mathrulezz ?

Answer 9

Yeah, chain rule is the way to go, lets look at it,
g(x) = sec(2x)
g'(x) = sec(2x)tan(2x)*(2x)'

Answer 10

is someone taking calculus 2?

Answer 11

Derivative of secx = secxtanx

Answer 12

And yes, C looks correct.

Answer 13

ok thank u guys :)

Answer 14

I also used prime notation, even though the one Jhanny showed you is the "legit" way, but this is easier to see what's going on especially if you're just starting derivatives.