Q.Find the derivative of the following: $y = x^{\log x} + (\log x)^x$

Two related techniques handle relationships between $x$ and $y$ (or functions of $x$) that the ordinary explicit-function derivative table cannot differentiate directly.

Implicit differentiation. An equation $F(x,y)=0$ (e.g. the circle $x^2+y^2=4$) may define $y$ as one or more functions of $x$ implicitly — without being solved for $y$ — and sometimes cannot even be solved for $y$ in elementary terms (e.g. $x^4+x^2y^3-y^5=2x+1$). The method: differentiate both sides of the equation with respect to $x$, treating $y$ throughout as a differentiable function of $x$ (so any term in $y$ picks up a factor $\dfrac{dy}{dx}$ via the chain rule — most simply as $\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}$), then solve algebraically for $\dfrac{dy}{dx}$. Worked example: for $x^2+y^2=1$, $2x+2y\,y'=0 \Rightarrow y'=-x/y$.

Logarithmic differentiation. Handles power-exponential functions such as $y=x^x$, where both the base and the exponent depend on $x$ and neither the plain power rule nor the plain exponential rule alone applies. Method: (1) take $\log$ of both sides and simplify with the log laws; (2) differentiate implicitly; (3) solve for $y'$. For $y=x^x$: $\log y=x\log x \Rightarrow \dfrac{y'}{y}=\log x+1 \Rightarrow y'=x^x(1+\log x)$. The technique covers four general cases — constant$^{\text{constant}}$ (derivative $0$), variable$^{\text{constant}}$ (ordinary power rule), constant$^{\text{variable}}$ (ordinary exponential rule), and variable$^{\text{variable}}$ (genuinely needs logs): $\dfrac{d}{dx}\big[f(x)^{g(x)}\big] = f(x)^{g(x)}\left[g'(x)\log f(x)+g(x)\dfrac{f'(x)}{f(x)}\right]$. It is also useful more broadly — even for a function with no variable exponent — whenever it is built from several products/quotients/powers, since $\log$ turns those into sums/differences/multiples before differentiating, which is often far less work than a direct product-and-quotient-rule computation.

A related shortcut — the substitution method. Some inverse-trigonometric expressions collapse dramatically under a trigonometric substitution before differentiating: e.g. for $f(x)=\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$, substituting $x=\tan\theta$ turns the argument into $\tan2\theta$ (a double-angle identity), so $f(x)=2\tan^{-1}x$, whose derivative $\dfrac{2}{1+x^2}$ is read off instantly — far simpler than differentiating the original quotient directly via the chain and quotient rules. The pattern to look for is a multiple-angle identity (double-angle, triple-angle, or tangent-addition) disguised inside the inverse trig function.