Sunday, May 19, 2019

IGNOU SOLVED ASSIGNMENT 2019 MTE01 1(a)(ii)

Given,
y = {\sinh ^{ - 1}}\left( {\tanh x} \right)

\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 + {{\tanh }^2}x} }}\frac{{d\tanh x}}{{dx}}
\frac{{dy}}{{dx}} = \frac{{{\rm{sec}}{{\rm{h}}^2}x}}{{\sqrt {1 + {{\tanh }^2}x} }}
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