Differentiate y = (sqrt((1−x)/(1+x))) with respect to x. Find f'(x) for f(x) = e3x (5x2 - 2x + 1).

Then $dx = -2\sin(2\theta)\,d\theta$.
Substitute $x$ into the expression inside the square root:

$$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}$$

$$\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \sqrt{\tan^2\theta} = |\tan\theta|$$

Assuming $x \in (-1, 1)$, then $2\theta \in (0, \pi)$, which means $\theta \in (0, \pi/2)$. In this interval, $\tan\theta > 0$.
So, $\sqrt{\frac{1-x}{1+x}} = \tan\theta$.

$$y = \arctan(\tan\theta)$$ $$y = \theta$$

Since $x = \cos(2\theta)$, we have $2\theta = \arccos x$, so $\theta = \frac{1}{2}\arccos x$.

$$y = \frac{1}{2}\arccos x$$

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\arccos x\right)$$ $$\frac{dy}{dx} = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$$ $$\boxed{\frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}}}$$

: Find $f'(x)$ for $f(x) = e^{3x} \ln(5x^2 - 2x + 1)$.

We use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.
Let $u = e^{3x}$ and $v = \ln(5x^2 - 2x + 1)$.

$$f'(x) = u'v + uv'$$ $$f'(x) = (3e^{3x})\ln(5x^2 - 2x + 1) + (e^{3x})\left(\frac{10x - 2}{5x^2 - 2x + 1}\right)$$

$$\boxed{f'(x) = e^{3x}\left(3\ln(5x^2 - 2x + 1) + \frac{10x - 2}{5x^2 - 2x + 1}\right)}$$

: Find $\frac{dy}{dx}$ if $y = \frac{\sin(4x^2)}{(x+1)^3}$.

We use the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.
Let $u = \sin(4x^2)$ and $v = (x+1)^3$.

$$\frac{dy}{dx} = \frac{(8x \cos(4x^2))(x+1)^3 - (\sin(4x^2))(3(x+1)^2)}{((x+1)^3)^2}$$ $$\frac{dy}{dx} = \frac{(8x \cos(4x^2))(x+1)^3 - 3\sin(4x^2)(x+1)^2}{(x+1)^6}$$

$$\frac{dy}{dx} = \frac{(x+1)^2 [8x \cos(4x^2)(x+1) - 3\sin(4x^2)]}{(x+1)^6}$$ $$\frac{dy}{dx} = \frac{8x(x+1)\cos(4x^2) - 3\sin(4x^2)}{(x+1)^4}$$ $$\boxed{\frac{dy}{dx} = \frac{8x^2 \cos(4x^2) + 8x \cos(4x^2) - 3\sin(4x^2)}{(x+1)^4}}$$

: Find $\frac{dy}{dx}$ if $y = x^x$.

We use logarithmic differentiation.

$$\ln y = \ln(x^x)$$

Use the logarithm property $\ln(a^b) = b \ln a$:

$$\ln y = x \ln x$$

Use the product rule on the right side.

$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x)$$ $$\frac{1}{y} \frac{dy}{dx} = (1)\ln x + x\left(\frac{1}{x}\right)$$ $$\frac{1}{y} \frac{dy}{dx} = \ln x + 1$$

$$\frac{dy}{dx} = y(\ln x + 1)$$

Substitute $y = x^x$ back into the equation:

$$\boxed{\frac{dy}{dx} = x^x(\ln x + 1)}$$

: Differentiate $y = \ln(\sec x + \tan x)^n$ with respect to $x$.

Recall that $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$.

$$\frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x$$

Factor out $\sec x$:

$$\sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$$

$$\frac{dy}{dx} = n \cdot \frac{1}{\sec x + \tan x} \cdot \sec x (\tan x + \sec x)$$

The term $(\sec x + \tan x)$ cancels out.

$$\boxed{\frac{dy}{dx} = n \sec x}$$