Solution:

Using the constant multiple rule and the derivative of the natural logarithm:

\[f'(x) = 7 \cdot \frac{1}{x} = \frac{7}{x}\]

Solution:

We need to apply the chain rule. Let \(u = g(x)=x^2 + 4x + 1\), and so \(y=f(u) = \ln(u)\). The chain rule states that \(y'=\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

Note that \(\frac{dy}{du}=\frac{d}{du}[\ln(u)] = \frac{1}{u}\) and \(\frac{du}{dx} = 2x + 4\).

Therefore:

\[f'(x) = \frac{1}{x^2 + 4x + 1} \cdot (2x + 4) = \frac{2x + 4}{x^2 + 4x + 1}\]

Solution:

We need to use the product rule and the chain rule. Let \(h(x) = (5x^2 - x)^2\) and \(k(x) = \sqrt[3]{3x} = (3x)^{1/3}\).

The product rule states that \(f'(x) = h'(x) \cdot k(x) + h(x) \cdot k'(x)\).

For \(h'(x)\), we use the chain rule:

\[h'(x) = 2(5x^2 - x) \cdot (10x - 1)\]

For \(k'(x)\), we also use the chain rule:

\[k'(x) = \frac{1}{3}(3x)^{-2/3} \cdot 3 = (3x)^{-2/3}\]

Combining these results:

\[f'(x) = 2(5x^2 - x)(10x - 1)\sqrt[3]{3x} + (5x^2 - x)^2(3x)^{-2/3}\]

This can be simplified as:

\[f'(x) = (100x^3 - 30x^2 + 2x)\sqrt[3]{3x} + \frac{(5x^2 - x)^2}{(3x)^{2/3}}\]

Solution:

We need to apply the chain rule twice. First, recognize that \(y = 3[\tan(3x)]^3\).

Let \(u = g(x)= \tan(3x)\), and so \(y=f(u) = 3u^3\). Hence, \(\frac{dy}{du} = 9u^2\).

Now we need \(\frac{du}{dx}\). Since \(u = \tan(3x)\), applying the chain rule yields:

\[\frac{du}{dx} = \sec^2(3x) \cdot 3 = 3\sec^2(3x)\]

Combining these results (and using the chain rule):

\[y'=\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}=9u^2\cdot3\sec^2(3x)= 9[\tan(3x)]^2\cdot 3\sec^2(3x) = 27\tan^2(3x)\sec^2(3x)\]

Solution:

We need to use both the product rule and the chain rule. Let \(h(x) = e^{ex}\) and \(g(x) = \sin(9x)\). The product rule states that \(f'(x) = h'(x) \cdot g(x) + h(x) \cdot g'(x)\). For \(h'(x)\), we apply the chain rule. Since \(h(x) = e^{ex}\), the outer function is \(e^u\) where \(u = ex\):

\[h'(x) = e^{ex} \cdot e=e^{ex+1}\]

For \(g'(x)\), we apply the chain rule with the sine function:

\[g'(x) = \cos(9x) \cdot 9 = 9\cos(9x)\]

Therefore:

\[f'(x) = e^{ex+1} \cdot \sin(9x) + e^{ex} \cdot 9\cos(9x)\]

Factoring out \(e^{ex}\):

\[f'(x) = e^{ex}[e\sin(9x) + 9\cos(9x)]\]

Solution:

We can simplify this function using logarithm properties before differentiating. First, note that the square root can be written as a power of one-half:

\[f(x) = \ln\left[\left(\frac{2x + 4}{x^2 - 4}\right)^{1/2}\right]\]

Using the logarithm property that \(\ln(a^b) = b\ln(a)\):

\[f(x) = \frac{1}{2}\ln\left(\frac{2x + 4}{x^2 - 4}\right)=\frac{1}{2}\ln\left(\frac{2(x + 2)}{(x+2)(x-2)}\right)=\frac{1}{2}\ln\left(\frac{2}{x-2}\right)\]

Using the logarithm property that \(\ln(a/b) = \ln(a) - \ln(b)\):

\[f(x) = \frac{1}{2}[\ln(2) - \ln(x-2)]\]

Now we can differentiate term by term:

\[f'(x) = \frac{1}{2}\left(0 - \frac{1}{x-2}\right)=\frac{1}{4-2x}\]