Question

derivative of log (sec x + than x)
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Answer 1

Let, y = ln(secx + tanx)

or, dy/dx ={1/(secx + tanx)} *{d/dx (secx + tanx)} or, dy/dx = (secx*tanx + sec^2x)/(secx + tanx)

or, dy/dx = secx(secx + tanx)/(secx + tanx)

or, dy/dx = secx ,

Therefore , derivative of ln(secx+tanx) is secx