Math Help - check :)

1. check :)

$
y = (1 + 4e^{2x})^{2x}
$

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

$
y = \left( \frac{1}{x} \right)^{ln x}
$

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

$
y = \frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3}
$

$
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)
$

$
y = \frac {ln x}{e^{\sqrt {x}}cos^4x}
$

$
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)
$

2. anything?

3. Hello, qzno!

$y \:=\: (1 + 4e^{2x})^{2x}$

$y' \:= \:(1 + 4e^{2x})^{2x} \left( \frac{16xe^{2x} + 8e^{2x}(1 + 4e^{2x})\ln(1 + 4e^{2x})}{(1 + 4e^{2x})}\right)$
I have a "2" where you have "8".

$y \:= \:\left(\frac{1}{x}\right)^{\ln x}$

I did it differently and got an entirely different-looking answer.

Watch this . . . $y \;=\;\left(x^{-1}\right)^{\ln x} \;=\;(x)^{-\ln x}$

Take logs: . $\ln(y) \;=\;\ln(x)^{-\ln x} \;=\;-\ln x\cdot\ln x \;=\;-(\ln x)^2$

Then: . $\frac{1}{y}\cdot y' \;=\;-2\cdot\ln x \cdot \frac{1}{x} \;=\;\frac{-2\ln x}{x}$

Therefore: . $y' \;=\;\left(\frac{1}{x}\right)^{\ln x}\left(\frac{-2\ln x}{x}\right)$

In the last two, be careful with your algebra.
. . You missed a "minus" in both of them.

$y \:= \:\frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3}$
We have: . $y \;=\;\frac{(x^3+1)^{\frac{1}{3}}}{x^3(x-1)^3}$

Take logs: . $\ln(y) \;=\;\ln\left[\frac{(x^3+1)^{\frac{1}{3}}}{x^3(x-1)^3}\right]$

. . $=\;\ln(x^3+1)^{\frac{1}{3}} - \ln\left[x^3(x-1)^3\right]$

. . $=\;\ln(x^3+1)^{\frac{1}{3}} - \underbrace{{\color{red}\bigg[}\ln(x^3) + \ln(x-1)^3{\color{red}\bigg]}}_{\text{minus }both\text{ of them!}}$

. . $= \;\ln(x^3+1)^{\frac{1}{3}} - \ln(x^3)\: {\color{red}-} \ln(x-1)^3$

Got it?

4. hahah thanks! ive been doing these for soo long today its not funny
is that last one right?