Derivative of underroot 1+sinxby 1-sinx
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Posted by Nishant Tiwari 7 years, 6 months ago
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Susai Raj 7 years, 6 months ago
Multiply both numerator and denominator with the conjucate of the denominator.
y=√{(1+sinx)(1+sinx) / (1-sinx)(1+sinx)}
y =√{(1+sinx)^2/(1-sin^2 x)
y =√{(1+sinx)2/cos^2 x}
y =(1+sinx)/cosx
Y =secx + tanx
So dy/dx = secx.tanx+sec^2 x
=secx(secx+tanx)
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