612.

$\mathrm{Find} \frac{\mathrm{dy}}{\mathrm{dx}}, \mathrm{if} \mathrm{y} = {\left( \mathrm{cosx}\right)}^{\mathrm{x}} + {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}}$

$$\begin{gathered} \mathrm{y} = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} + {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \\ \mathrm{For} \mathrm{simplication}, \mathrm{Let} \mathrm{us} \mathrm{consider} \mathrm{y} = \mathrm{A} + \mathrm{B} \mathrm{such} \mathrm{that} \mathrm{A} = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \\ \mathrm{and} \mathrm{B} = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \\ \mathrm{Then}, \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dA}}{\mathrm{dx}} + \frac{\mathrm{dB}}{\mathrm{dx}} .........\left(\mathrm{i}\right) \\ \mathrm{A} = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \end{gathered}$$

Taking logarithms on both the sides, we have,

log A = x log ( cosx )

$$\begin{gathered} \frac{1}{\mathrm{A}}\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{\mathrm{d}}{\mathrm{dx}} \left[ \mathrm{x} \log \left(\mathrm{cosx} \right)\right] \\ \Rightarrow \frac{{\displaystyle \mathrm{dA}}}{{\displaystyle \mathrm{dx}}} = \mathrm{A}\frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{dx}}} \left[ \mathrm{x} \log \left(\mathrm{cosx} \right)\right] \\ = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \left[ \mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}} \left[ \log \left( \mathrm{cosx} \right)\right] +\log \left(\mathrm{cosx} \right) \frac{\mathrm{d}}{\mathrm{dx}} \left( \mathrm{x} \right) \right] \\ = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \left[ \mathrm{x}\frac{1}{\cos {\displaystyle \mathrm{x}}} \left( - \mathrm{sinx} \right) + \log \left( \mathrm{cosx} \right) \left( 1 \right)\right] \\ = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \left[ - \mathrm{x} \mathrm{tanx} + \log \left( \mathrm{cosx} \right)\right] ........\left(\mathrm{ii}\right) \\ \mathrm{Now}, \mathrm{B} = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \\ \mathrm{B} = {\left( \mathrm{sinx} \right)}^{\frac{{\displaystyle 1}}{{\displaystyle \mathrm{x}}}} \end{gathered}$$

Taking logarithms on both the sides, we have,

$$\begin{gathered} \mathrm{Log} \mathrm{B} = \frac{1}{\mathrm{x}} \log \left( \mathrm{sinx} \right) \\ \frac{1}{\mathrm{B}}\frac{\mathrm{dB}}{\mathrm{dx}} = \frac{\mathrm{d}}{\mathrm{dx}} \left[ \frac{1}{\mathrm{x}} \log \left( \mathrm{sinx} \right) \right] \\ \Rightarrow \frac{\mathrm{dB}}{\mathrm{dx}} = \mathrm{B}\frac{\mathrm{d}}{\mathrm{dx}} \left[ \frac{1}{\mathrm{x}} \log \left( \mathrm{sinx} \right) \right] \\ = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \left[\frac{1}{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}} \left( \log \left( \mathrm{sinx} \right)\right)+\log \left( \mathrm{sinx} \right) \frac{\mathrm{d}}{\mathrm{dx}} \left( \frac{1}{\mathrm{x}}\right) \right] \\ = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \left[\frac{1}{\mathrm{x}}\frac{1}{\mathrm{sinx}} \left( \mathrm{cosx} \right)+\log \left( \mathrm{sinx} \right) \left(-\frac{1}{{\mathrm{x}}^2}\right) \right] \\ = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \left[ \frac{1}{\mathrm{x}} \mathrm{cotx} - \left(-\frac{1}{{\mathrm{x}}^2}\right) \log \left( \mathrm{sinx} \right) \right] \\ = {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \left[ \frac{\mathrm{x} \mathrm{cotx} -\log \left( \mathrm{sinx} \right) }{{\mathrm{x}}^2}\right] .......\left(\mathrm{iii}\right) \end{gathered}$$

Now, on substituting  (ii) and (iii) in (i) we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left( \mathrm{cosx} \right)}^{\mathrm{x}} \left[ - \mathrm{x} \mathrm{tanx} + \log \left( \mathrm{cosx} \right)\right] + {\left( \mathrm{sinx} \right)}^{\frac{1}{\mathrm{x}}} \left[ \frac{ \mathrm{x} \mathrm{cotx} - \log \left( \mathrm{sinx} \right)}{{\mathrm{x}}^2}\right]$