If \[y = \sin ({x^x})\], prove that \[\dfrac{{dy}}{{dx}} = \cos ({x^x}).{x^x}(1 + \log x)\]

Hint: We assume the angle inside the function of sine as a variable and find the derivative of y with respect to that variable. Now we differentiate the equation where we assumed the angle as a variable. Take log on both sides of the equation and use the property of log to break it. Use the product rule of differentiation to differentiate it. Substitute the value of differentiation of variable with respect to x in the equation where we had differentiation of y with respect to the variable.
* \[\log {m^n} = n\log m\]
* Product rule of differentiation:\[\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)\]
* Chain rule of differentiation:\[\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)\]

Complete step-by-step answer:
We are given that \[y = \sin ({x^x})\]
Let us assume that \[u = {x^x}\] … (1)
Then the value of y becomes \[y = \sin u\]
Differentiate both sides of equation with respect to x
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\sin u} \right)\]
We know differentiation of sine is cosine
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos u.\dfrac{{du}}{{dx}}\] … (2)
Now we have from equation (1) \[u = {x^x}\]
We take log on both sides of the equation
\[ \Rightarrow \log u = \log {x^x}\]
Use the property of log on right side of the equation i.e. \[\log {m^n} = n\log m\]
\[ \Rightarrow \log u = x\log x\]
Differentiate both sides of the equation with respect to x
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log u} \right) = \dfrac{d}{{dx}}\left( {x(\log x)} \right)\]
Apply chain rule of differentiation in LHS of the equation
Chain rule gives us\[\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)\].
Here\[g(f(x)) = \log (u),f(x) = u\], then the equation becomes
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log u} \right) \times \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {x(\log x)} \right)\]
We know \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]
\[ \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {x(\log x)} \right)\]
Now apply product rule of differentiation in RHS of the equation
Product rule gives us\[\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)\]
Here\[a = x,b = \log x\], then the equation becomes
\[ \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x\]
Substitute the values\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]and\[\dfrac{{dx}}{{dx}} = 1\]in RHS of the equation
\[ \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = x \times \dfrac{1}{x} + \log x \times 1\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \dfrac{1}{u} \times \dfrac{{du}}{{dx}} = \left( {1 + \log x} \right)\]
Shift or cross multiply value of u from LHS to RHS
\[ \Rightarrow \dfrac{{du}}{{dx}} = u\left( {1 + \log x} \right)\] … (3)
Substitute this value from equation (3) in equation (2)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos u.u\left( {1 + \log x} \right)\]
Substitute the value of u from equation (1)
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos ({x^x}).{x^x}\left( {1 + \log x} \right)\]

Hence Proved

Note:
Many students make the mistake of calculating the differentiation of y with respect to x and leave it in terms of other variables which is wrong, if we use any substitution we have to substitute the value of differentiation of that substitution back as well.