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Mathematics 6 Online
OpenStudy (luigi0210):

Derivatives question:

OpenStudy (luigi0210):

Does: \[\LARGE \frac{d}{dx}sec^2x=\frac{d}{dx}tan^2x\]

OpenStudy (anonymous):

How did you get that?

OpenStudy (anonymous):

If it were true, that would mean that: \[ \sec^2(x) - \tan^2(x) = C \]

OpenStudy (gorv):

no.....dtan^2x/dx=2tanx*sec^2x

OpenStudy (gorv):

sorry yes

OpenStudy (anonymous):

Did you use the Pythagorean property, or did you directly differentiate?

OpenStudy (luigi0210):

It was a quiz question, I was curious.

OpenStudy (gorv):

\[\sec ^2x -\tan ^2x=1\]

OpenStudy (anonymous):

You could differentiate both sides of:\[ 1+\tan^2(x)=\sec^2(x) \]

OpenStudy (gorv):

differentianting \[\frac{ d(\sec^2x-\tan^2x) }{ dx }=\frac{ d (1) }{ dx}=0\]

OpenStudy (gorv):

\[\frac{ d \sec^2x }{ dx } - \frac{ d \tan^2x }{ dx }=0\]

OpenStudy (gorv):

@Luigi0210

OpenStudy (isaiah.feynman):

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