\displaystyle y=\ \ln \left( {\frac{{{{e}^{{{{x}^{2}}}}}}}{{\cos (1-x)}}} \right)

Before you Differentiate, simplify the expression

using the Laws of Logarithm   \displaystyle {{\log }_{a}}(\frac{x}{y})\ =\log {}_{a}x-{{\log }_{a}}y

\displaystyle y=\text{ln}\left( {{{e}^{{{{x}^{2}}}}}} \right)-\ln \left[ {\cos (1-x)\ } \right]

and Laws of Logarithm again  \displaystyle {{\log }_{a}}{{x}^{c}}\ =c\log {}_{a}x

\displaystyle y={{x}^{2}}\text{ln}\left( e \right)-\ln \left[ {\cos (1-x)\ } \right]

\displaystyle y={{x}^{2}}-\ln \left[ {\cos (1-x)\ } \right]

Differentiate the expression and simplify.

\displaystyle \frac{{dy}}{{dx}}=\frac{{{{x}^{3}}}}{3}-\frac{1}{{\cos (1-x)\ }}\times -\sin (1-x)\times -1

\displaystyle \frac{{dy}}{{dx}}=\frac{{{{x}^{3}}}}{3}-\frac{{\sin (1-x)}}{{\cos (1-x)\ }}

\displaystyle \frac{{dy}}{{dx}}=\frac{{{{x}^{3}}}}{3}-\tan (1-x)


You can differentiate without first simplifying, but you will find that you will take more steps to get the answer. REMEMBER: The More (unnecessary) Steps = The More Mistakes.  See Post about Differentiation of ln with sine


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