How to find derivative of function What is algorithm to find differentiation of function $x^{2^x}$? What formula I need to apply? Thanks.
 A: $$\large 
y  = x^{2^x}\\
\ln y  = 2^x \ln x\\
\dfrac{y'}y  =(\ln2) 2^x \ln x + \frac1x 2^x\\
y'  = y \left( (\ln2) 2^x \ln x + \frac1x 2^x \right) \\
$$

$$ \large
 y'  = x^{2^x} \left( (\ln2) 2^x \ln x + \frac1x 2^x \right) \\
$$

A: Say $y = x^{2^x}$.
Use logarithmic differentiation for this problem.
Taking $\ln$ on both sides,
$$\ln(y) = 2^x \cdot \ln(x)$$
Now, differentiate with respect to x on both sides
$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(2^x \cdot \ln(x))$$ (Use product rule here.)
Simplify the above, rearrange the terms to get your answer.
A: Just for fun:
Let 
$$  z=f(u,v)=u^v,$$
$$u=x,$$
and
$$v=2^x.$$ 
By the multivariable chain rule,
$$
{dz\over dx}={\partial z\over \partial u}{du\over dx}+{\partial z\over \partial v}{dv\over dx},
$$
we have
$$\eqalign{
{d\over dx } x^{2^x}={dz\over dx}
&={\partial\over\partial u} \,u^v\cdot{d\over dx}\, x+ {\partial\over \partial v}\, u^v\cdot{d\over dx}\, 2^x\cr
&=\strut\ v\, u^{v-1}\cdot1\ + \ u^v (\ln u)\cdot2^x(\ln 2)\cr
&=\strut2^x x^{2^x-1}+ x^{2^x} (\ln x)  \cdot (\ln 2 )2^x\cr
&=x^{2^x}\bigl(\textstyle{2^x\over x}+(\ln 2)(\ln x) 2^x\bigr).
}$$
A: To solve diff. of this type $f(x)^{\large g(x)}$ use log to easy this function $$y=x^{\large 2^{\large x}}$$
$$\log y=\large 2^{\large x}\large\log  x$$
diff.with respect to x,using property of multiplication of function:
 $$ \dfrac 1y\cdot\dfrac {dy}{dx}=\large 2^{\large x}\cdot \dfrac 1x+\large\log  x\cdot \large 2^{\large x}\cdot\log2$$
 $$ \dfrac {dy}{dx}=\large y\left(\large 2^{\large x}\cdot \dfrac 1x+\large\log  x\cdot \large 2^{\large x}\cdot\log2\right)$$
 $$ \dfrac {dy}{dx}=\large x^{\large 2^{\large x}}\large \left(\large 2^{\large x}\cdot \dfrac 1x+\large\log  x\cdot \large 2^{\large x}\cdot\log2\right)$$
