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Derivatives question:
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Does: \[\LARGE \frac{d}{dx}sec^2x=\frac{d}{dx}tan^2x\]
How did you get that?
If it were true, that would mean that: \[ \sec^2(x) - \tan^2(x) = C \]
no.....dtan^2x/dx=2tanx*sec^2x
sorry yes
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Did you use the Pythagorean property, or did you directly differentiate?
It was a quiz question, I was curious.
\[\sec ^2x -\tan ^2x=1\]
You could differentiate both sides of:\[ 1+\tan^2(x)=\sec^2(x) \]
differentianting \[\frac{ d(\sec^2x-\tan^2x) }{ dx }=\frac{ d (1) }{ dx}=0\]
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\[\frac{ d \sec^2x }{ dx } - \frac{ d \tan^2x }{ dx }=0\]
@Luigi0210
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