Question PLeASE FIND THE TANGENT LINE AND THE NORMAL LINE FOR POINT \( (-1,0) \) ON THE CURVE GIVEN BY: \[ y^{3} \cosh x=e^{y}+x \]
【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2We find equation of the tangent line and normalization of the curve at the given pointExplanation:Please refer to solution in this step.Step2/2Given \( \mathrm{{y}^{{3}}{\cosh{{x}}}={e}^{{y}}+{x}} \)Differentiate with respect to \( \mathrm{'{x}',} \) We get\( \begin{align*} \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left[{y}^{{3}}{\cosh{{x}}}\right]}} &= \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left[{e}^{{y}}+{x}\right]}}\\[3pt]\mathrm{\Rightarrow{3}{y}^{{2}}{\cosh{{x}}}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{y}^{{3}}{\sinh{{x}}}} &= \mathrm{{e}^{{y}}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{1}}\\[3pt]\mathrm{\Rightarrow{\left({3}{y}^{{2}}{\cosh{{x}}}-{e}^{{y}}\right)}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} &= \mathrm{{1}-{y}^{{3}}{\sinh{{x}}}}\\[3pt]\mathrm{\Rightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} &= \mathrm{\frac{{{1}-{y}^{{3}}{\sinh{{x}}}}}{{{3}{y}^{{2}}{\cosh{{x}}}-{e}^{{y}}}}\ \text{ }\ \ \text{ }\ \leftarrow{s}{l}{o}{p}{e}} \end{align*} \)At \( \mathrm{{\left(-{1},{0}\right)}} \)\( \begin{align*} \mathrm{{m}=\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}} &= \mathrm{\frac{{{1}-{0}.{\sinh{{\left(-{1}\right)}}}}}{{{ ... See the full answer