$\displaystyle y = x^{x^3}$
$\displaystyle y' = x^{2x^3} (1 - 3 ln x)$

$\displaystyle y = (cos 2x)^{\sqrt{x}}$
$\displaystyle y' = (cos 2x)^{\sqrt{x}} \left ( \frac{cos2xlncos2x - 4xsin2x}{2cos2x \sqrt{x}} \right)$

2. Hello,
Originally Posted by qzno
$\displaystyle y = x^{x^3}$
$\displaystyle y' = x^{2x^3} (1 - 3 ln x)$
There are some problems.
First of all, $\displaystyle a^b a^c=a^{b+c}$ and not $\displaystyle a^{bc}$

$\displaystyle y=e^{x^3 \ln(x)}$
So :
$\displaystyle y'=(x^3 \ln(x))' e^{x^3 \ln(x)}=x^{x^3} (x^3 \ln(x))'$

Use product rule to calculate $\displaystyle (x^3 \ln(x))'$ :
$\displaystyle =3x^2 \ln(x)+x^3 \cdot \frac 1x=3x^2 \ln(x)+x^2=x^2(3 \ln(x)+1)$

So $\displaystyle y'=x^{x^3} \cdot x^2(3 \ln(x)+1)=x^{\color{red}x^3+2} (3 \ln(x)+1)$

Originally Posted by qzno

$\displaystyle y = (cos 2x)^{\sqrt{x}}$
$\displaystyle y' = (\cos (2x))^{\sqrt{x}} ( \frac{\cos(2x) \ln(\cos(2x)) - 4x \sin(2x)}{2\cos(2x) \sqrt{x}})$
I changed some parts of the code

Anyway, the last one looks perfect to me. Congratulations

3. hahah thanks! do you mind checking some more? if not:

$\displaystyle y = (1 + 4e^{2x})^{2x}$
$\displaystyle y' = (1 + 4e^{2x})^{2x} \left( \frac{16xe^{2x} + 8e^{2x}(1 + 4e^{2x})ln(1 + 4e^{2x})}{(1 + 4e^{2x})}\right)$

$\displaystyle y = \left( \frac{1}{x} \right)^{ln x}$
$\displaystyle y' = \left( \frac{1}{x} \right)^{ln x} \left(\frac{x^2lnx + ln \frac{1}{x}}{x}\right)$

$\displaystyle y = \frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3}$
$\displaystyle y' = \frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3} \left( \frac{x^2}{x^3+1} - \frac{3}{x} + \frac{3}{x-1} \right)$

$\displaystyle y = \frac {ln x}{e^{\sqrt {x}}cos^4x}$
$\displaystyle y' = \frac {ln x}{e^{\sqrt {x}}cos^4x} \left( \frac{1}{xlnx} - \frac{e^{\sqrt{x}}}{2 \sqrt{x} e^{\sqrt{x}}} - 4tanx \right)$