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 :)
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OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
i think its c
OpenStudy (jhannybean):
This involves using the chain rule: \(\frac{d}{dx}(\sec(x) \cdot \frac{d}{dx}(x)\)
OpenStudy (anonymous):
2sec(2x)tan(2x)
OpenStudy (anonymous):
c is the answer
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OpenStudy (anonymous):
that's what i got too @laetitiacarmel :) and @Jhannybean ok, do u think its correct as C? :)
OpenStudy (anonymous):
yes
OpenStudy (jhannybean):
Do you know your derivatives @mathrulezz ?
OpenStudy (anonymous):
Yeah, chain rule is the way to go, lets look at it,
g(x) = sec(2x)
g'(x) = sec(2x)tan(2x)*(2x)'
OpenStudy (anonymous):
is someone taking calculus 2?
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OpenStudy (anonymous):
Derivative of secx = secxtanx
OpenStudy (jhannybean):
And yes, C looks correct.
OpenStudy (anonymous):
ok thank u guys :)
OpenStudy (anonymous):
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.