How to find derivative of function

Question

What is algorithm to find differentiation of function $x^{2^x}$? What formula I need to apply? Thanks.

Answer 1

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

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

Answer 2

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 $$\eqalign{ {d\over dx } x^{2^x}={dz\over dx} &={\partial\over\partial u} \,u^v\cdot{d\over dx}\, x+ {\partial\over \partial v}\, u^v\cdot{d\over dx}\, 2^x\cr &=\strut\ v\, u^{v-1}\cdot1\ + \ u^v (\ln u)\cdot2^x(\ln 2)\cr &=\strut2^x x^{2^x-1}+ x^{2^x} (\ln x) \cdot (\ln 2 )2^x\cr &=x^{2^x}\bigl(\textstyle{2^x\over x}+(\ln 2)(\ln x) 2^x\bigr). }$$

Answer 3

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.

Answer 4

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: $$ \dfrac 1y\cdot\dfrac {dy}{dx}=\large 2^{\large x}\cdot \dfrac 1x+\large\log x\cdot \large 2^{\large x}\cdot\log2$$ $$ \dfrac {dy}{dx}=\large y\left(\large 2^{\large x}\cdot \dfrac 1x+\large\log x\cdot \large 2^{\large x}\cdot\log2\right)$$ $$ \dfrac {dy}{dx}=\large x^{\large 2^{\large x}}\large \left(\large 2^{\large x}\cdot \dfrac 1x+\large\log x\cdot \large 2^{\large x}\cdot\log2\right)$$