amyna
  • amyna
y=-3(sec^2x - tan^2(x)) + 4x^3 I do not know how to solve this! Please help with the calc problem!! Thank you!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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dan815
  • dan815
sec(x) = 1/cos(x)
freckles
  • freckles
recall Pythagorean identities \[\sin^2(x)+\cos^2(x)=1 \\ \text{ divide both sides by } \cos^2(x) \\ \tan^2(x)+1=\sec^2(x) \\\] anyways I don't know what it is meant by solve
freckles
  • freckles
could you say what solve means here @amyna if you are here

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freckles
  • freckles
you say this is a cal problem do you really mean differentiate are you trying to find the inverse like what are you doing exactly
amyna
  • amyna
yup sorry! differentiate it, thanks!
freckles
  • freckles
so you need help to differentiate -3+4x^3 just use constant rule and constant multiple rule and power rule?
freckles
  • freckles
-3 is a constant the derivative of a constant is 0
freckles
  • freckles
\[\frac{d}{dx} cx^n \\ c \frac{d}{dx} x^n \text{ by constant multiple rule } \\ c n x^{n-1} \text{ by power rule }\]
freckles
  • freckles
\[\frac{d}{dx}(-3)=? \\ \frac{d}{dx}4x^3=?\]
amyna
  • amyna
|dw:1436817831485:dw|
freckles
  • freckles
where did all the sec and tan stuff come from
freckles
  • freckles
did you use the Pythagorean identity earlier ? you should y=-3+4x^3
freckles
  • freckles
and we are wanting to find y' now
freckles
  • freckles
y'=0+12x^2=12x^2
freckles
  • freckles
though if you really want to differentiate sec^2(x)-tan^2(x) which is a constant so the derivative should be 0 but pretending you don't know you should wind up with this: \[\frac{d}{dx} \sec^2(x)=2 \sec(x) \cdot \sec(x)\tan(x) =2\sec^2(x)\tan(x) \\ \frac{d}{dx} \tan^2(x)=2 \tan(x) \sec^2(x) =2 \sec^2(x)\tan(x) \\ \text{ but as you notice } \\ \frac{d}{dx} \sec^2(x)-\frac{d}{dx} \tan^2(x)=0 \\ \frac{d}{dx}(\sec^2(x)-\tan^2(x))=0\] this is because \[\sec^2(x)-\tan^2(x) \text{ is a constant } \\ \text{ this is because } \sec^2(x)-\tan^2(x)=1 \text{ for all } x \text{ on it's domain }\]
freckles
  • freckles
but as I said all of that work is unnecessary
freckles
  • freckles
your y really just say y=-3+4x^3
amyna
  • amyna
so what is the final answer?
freckles
  • freckles
do you know how to differentiate -3 and 4x^3?
amyna
  • amyna
yes its 0 and 12x^2
freckles
  • freckles
ok great then why are you asking me what y' is when you know \[y=-3+4x^3 \\ y'=0+12x^2 \\ y'=12x^2 \]
amyna
  • amyna
oh okay! i was looking at it wrong! now i get it! thank you so much!

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