Sunday, May 19, 2019

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

Given,

y = {x^x} + {\left( {\sin x} \right)^{\ln x}}

First,

\begin{array}{c} {y_1} = {x^x}\\ \ln {y_1} = x*\ln x\\ \frac{1}{{{y_1}}}\frac{{d{y_1}}}{{dx}} = \left[ {1 + \ln x} \right]\\ \frac{{d{y_1}}}{{dx}} = {x^x}\left[ {1 + \ln x} \right] \end{array}

Now,

{y_2} = {\left( {\sin x} \right)^{\ln x}}
\ln {y_2} = \ln x*\ln \sin x
\frac{1}{{{y_2}}}\frac{{d{y_2}}}{{dx}} = \frac{{\ln \sin x}}{x} + \frac{{\ln x}}{{\sin x}}

\frac{{d{y_2}}}{{dx}} = {\left( {\sin x} \right)^{\ln x}}\left[ {\frac{{\ln \sin x}}{x} + \frac{{\ln x}}{{\sin x}}} \right]

Adding the derivatives of {y_1} and {y_2} we get

\frac{{dy}}{{dx}} = {x^x}\left[ {1 + \ln x} \right] + {\left( {\sin x} \right)^{\ln x}}\left[ {\frac{{\ln \sin x}}{x} + \frac{{\ln x}}{{\sin x}}} \right]

as the required answer.



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