Sunday, May 19, 2019

IGNOU MTE 01 2019 SOLVED ASSIGNMENT 1(b)

Put
x = \sec \theta
{f^'}\left( {\sec \theta } \right) = \sqrt {{{\sec }^2}\theta - 1}
{f^'}\left( {\sec \theta } \right) = \tan \theta
Integrating the above statement
f\left( {\sec \theta } \right) = - {{\rm logcos}\nolimits} \theta
f\left( {\sec \theta } \right) = \log \frac{1}{{\cos \theta }}
f\left( {\sec \theta } \right) = {{\rm logsec}\nolimits} \theta
f\left( x \right) = {{\rm logx}\nolimits}
Now,
f\left( {{x^2}} \right) = {{{\rm logx}\nolimits} ^2}
y = {{{\rm logx}\nolimits} ^2}
y = 2{{\rm logx}\nolimits}
\frac{{dy}}{{dx}} = \frac{2}{x}
Posted by at

1 comment:

Subscribe to: Post Comments (Atom)