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find the curvature of y=tanx using the formula |f''(x)|/[1+(f'(x))^2]^(3/2). I have (2tanxsec^2x)/[1+sec^4(x)]^(3/2) and not sure how to simplify
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Hold on so the curvature is given by: \[ \frac{|f''(x)|}{(1+[f'(x)]^2)^{3/2}} \]
Is that what you're doing?
How do you know you need to simplify?
Nevermind I got it
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