$$f(x) = (x^3 + x^2 - x + 1)^3$$

Inner function:   $x^3 + x^3 - x + 1$

Outer function:   $x^3$

$$f'(x) = 3(x^3 + x^3 - x + 1)(3x^2 + 2x - 1)$$

$$f(x) = \sqrt{1 - 3x}$$

Inner function:   $1 - x^3$

Outer function:   $x^{\frac{1}{2}}$

$$\begin{align} f'(x) &= \frac{1}{2} (1 - 3x)^{-\frac{1}{2}}(-3) \\ &= \frac{-3}{2 \sqrt{1 - 3x}} \end{align}$$

Note: in calculus, it's almost always best to replace radicals with rational exponents.

$$f(x) = \frac{1}{x^2 - 1}$$

First rewrite the function as $f(x) = (x^2 - 1)^{1/2}$

Inner function:   $x^2 - 1$

Outer function:  $x^{\frac{1}{2}}$

$$\begin{align} f'(x) &= -(x^2 - 1)^{-2} \cdot 2x \\ \\ &= \frac{-2x}{(x^2 - 1)^2} \end{align}$$

Note that this derivative could also be done by using the quotient rule directly. Try that and make sure you get the same derivative

$$f(x) = \frac{1}{1 + cos(x)}$$

First rewrite the function as $f(x) = (1 + cos(x))^{-1}$

Inner function:   $1 + cos(x)$

Outer function:  $x^{-1}$

$$\begin{align} f'(x) &= -(1 + cos(x))^{-2} \cdot -sin(x) \\ \\ &= \frac{sin(x)}{[1 + cos(x)]^2} \end{align}$$

Note that this derivative could also be done by using the quotient rule directly. Try that and make sure you get the same derivative

$$f(x) = \frac{1}{cos^2(x)}$$

If we rewrite f(x) as $f(x) = (cos^2(x))^{-1}$, and realize that the funky shorthand we use for raising trig functions to powers actually means

$$f(x) = [(cos(x))^2]^{-1},$$

Then our function, written in the best shape to use the chain rule, is $f(x) = [cos(x)]^{-2}$

Inner function:   $cos(x)$

Outer function:  $x^{-2}$

$$\begin{align} f'(x) &= -2 [cos(x)]^{-3} \cdot -sin(x) \\ \\ &= \frac{2 sin(x)}{cos^3(x)} \end{align}$$

This derivative could also be done by using the quotient rule directly. Try that and make sure you get the same result.

$$f(x) = x^2 + cos^2(x)$$

This is the sum of two functions; recall that the derivative of a sum is the sum of derivatives, so the derivative of the first term is just $2x$. The second function is a composite function:

Inner function:   $cos(x)$

Outer function:  $x^2$

$$\begin{align} f'(x) &= 2x + 2cos(x)\cdot (-sin(x)) \\ \\ &= 2[x - sin(x) cos(x)] \end{align}$$

$$f(\theta) = 3 cot (n\theta)$$

The inner function is very simple for this composite function $(3\theta)$, but it is there, so the chain rule is necessary.

Inner function:   $3 \theta$

Outer function:  $cot(x)$

$$f'(\theta) = 3 n\cdot csc^2(n\theta)$$

If you forgot the derivative of $f(x) = cot(x)$, it's easy to figure out just by rewriting it as $\frac{cos(x)}{sin(x)}$ and using the quotient rule.

$$f(x) = (2x - 3)^4(x^2 + x + 1)^4$$

This is a product of composite functions: $f(g(x)) \cdot f(h(x))$, where   $f(x) = x^4, \; g(x) = 2x - 2 \; \text{and} \; h(x) = x^2 + x + 1$

$$\begin{align} f'(x) &= 4(2x - 3)^3 \cdot 2 \cdot (x^2 + x + 1)^4 \\ &+ (2x - 3)^4 \cdot 4 \cdot (x^2 + x + 1)^3 (2x + 1) \\ \\ &= 4(x^2 + x + 1)^3\cdot(2x - 3)^3 \cdot \\ &[2(x^2 + x + 1) + (2x + 1)(2x - 3)] \end{align}$$

$$f(x) = \left(\frac{x^2 + 1}{x^2 - 1}\right)^3$$

Inner function:   $\frac{x^2 + 1}{x^2 - 1}$

Outer function:  $x^3$

After taking the simple derivative of the outer function, we'll have to multiply it by the quotient-rule derivative of the inner function.

$$\begin{align} f'(x) &= 3 \left(\frac{x^2 + 1}{x^2 - 1}\right)^2 \left(\frac{x^2 - 1)\cdot 2x - (x^2 + 1)(2x)}{(x^2 - 1)^2}\right) \\ \\ &= 3 \left(\frac{x^2 + 1}{x^2 - 1}\right)^2 \left( \frac{2x(x^2 - 1 - x^2 - 1)}{(x^2 - 1)^2}\right) \\ \\ &= 3 \left(\frac{x^2 + 1}{x^2 - 1}\right)^2 \left( \frac{-2x}{(x^2 - 1)^2} \right) \end{align}$$

Rewrite the radical as a rational exponent:

$$f(x) = \sqrt{\frac{x^2 + 1}{x^2 + 4}} = \left( \frac{x^2 + 1}{x^2 + 4}\right)^{\frac{1}{2}}$$

Inner function:   $\frac{x^2 + 1}{x^2 + 4}$

Outer function:  $x^{\frac{1}{2}}$

After taking the simple derivative of the outer function, we'll have to multiply it by the quotient-rule derivative of the inner function.

$$\begin{align} f'(x) &= 3 \left(\frac{x^2 + 1}{x^2 + 4}\right)^{-\frac{1}{2}} \left(\frac{x^2 + 4)\cdot 2x - (x^2 + 1)(2x)}{(x^2 + 4)^2}\right) \\ \\ &= \frac{1}{2} \left(\frac{x^2 + 1}{x^2 + 4}\right)^{-\frac{1}{2}} \left( \frac{2x(x^2 + 4 - x^2 - 1)}{(x^2 + 4)^2}\right) \\ \\ &= \frac{1}{2} \left(\frac{x^2 + 1}{x^2 + 4}\right)^{-\frac{1}{2}} \left( \frac{6x}{(x^2 + 4)^2} \right) \end{align}$$

This could be simplified more, of course ...

It can help to rewrite this function as a product, and also use a rational exponent instead of the radical sign:

$$f(x) = \frac{x}{\sqrt{7 - 3x}} = x (7 - 3x)^{-\frac{1}{2}}$$

Inner function:   $7 - 3x$

Outer function:   $x^{-\frac{1}{2}}$

Now use the product rule and the chain rule:

$$\begin{align} f'(x) &= (1)(7 - 3x)^{-\frac{1}{2}} - x \left(\frac{1}{2}\right)(7 - 3x)^{-\frac{3}{2}}(3) \\ \\ &= \frac{1}{\sqrt{7 - 3x}} - \frac{3x}{2(7 - 3x)^{\frac{3}{2}}} \end{align}$$

$$f(x) = \frac{(x - 1)^4}{(x^2 + 2x)^5}$$

Let's do this one using the quotient rule, but we'll need the chain rule to do the inner derivatives:

$$\begin{align} f'(x) &= \frac{(x^2 + 2x)^5 \cdot 4(x - 1)^3 - (x - 1)^4 (5) (x^2 + 2x)^4(2x + 2)}{(x^2 + 2x)^{10}} \\ \\ &= \frac{(x^2 + 2x)^4 \cdot (x - 1)^3 [4(x^2 + 2x) - 5(x - 1)(2x + 2)]}{(x^2 + 2x)^{10}} \\ \\ &= \frac{(x - 1)^3[4x^2 + 8x - 10x^2 + 10]}{(x^2 + 2x)^{6}} \\ \\ &= \frac{(x - 1)^3[-6x^2 + 8x + 10]}{(x^2 + 2x)^{6}} \end{align}$$

Of course, this could also be done by experssing the quotient as a product and using the product and chain rules.

$$f(x) = sin(cos(2x))$$

This function is a composite of three functions $f(g(h(x)))$, where $h(x) = 2x$, $g(x) = cos(x)$ and $f(x) = sin(x)$. This is a good example of why we call this the chain rule: We take the derivative of the outermost function, leaving everything else as is, multiply (or "chain") the derivatie of the next inner function, leaving the innermost intact, and finally, multiply on to the growing chain the derivative of the innermost functions.

$$f'(x) = cos(cos(2x)) \cdot -sin(2x) (2)$$

$$f'(x) = -2 sin (2x) \cdot cos(cos(2x))$$

(The second rearrangement was just for style)

$$f(x) = \left(\frac{1 - cos(2x)}{1 + cos(2x)}\right)^4$$

Inner function:   $\frac{1 - cos(2x)}{1 + cos(2x)}$

Outer function:   $x^4$

The derivative of the inner function will have to be found using the quotient rule.

$$\begin{align} f'(x) &= 4 \left( \frac{1 - cos(2x)}{1 + cos(2x)} \right)^3 \cdot \frac{(1 + cos(2x))\cdot 2 sin(2x) - (1 - cos(2x))(-2 sin(2x))}{(1 + cos(2x))^2} \\ \\ &= 4 \left( \frac{1 - cos(2x)}{1 + cos(2x)} \right)^3 \cdot \frac{2\cdot sin(2x)[(1 + cos (2x)) + (1 - cos(2x))]}{(1 + cos(2x))^2} \end{align}$$

Quotient-rule derivatives are often difficult to simplify, with plenty of opportunity to make small errors. Don't let that take up too much of your time. Take care of the obvious simplifications and let it go.

$$f(x) = [x^2 + (1 - 3x)^5]^3$$

$$f'(x) = 3 [x^2 + (1 - 3x)^5]^2 \cdot [2x + 5(1 - 3x)^4 (-3)]$$

Notice that the last inner derivative (-3) applies only to the $(1 - 3x)^5$ term.

There's not much point in simplifying here. You might not really gain that much. That's the way it goes, sometimes.

Let's rewrite this function as a product using a -1 exponent on the denominator. Then we can use the product rule and chain rule.

$$f(x) = \frac{cos(x)}{sin(x) + cos(x)} = cos(x)[sin(x) + cos(x)]^{-1}$$

$$\begin{align} f'(x) &= -sin(x) (sin(x) + cos(x))^{-1} + cos(x)[-sin(x) + cos(x))^{-2}][cos(x) - sin(x)] \\ \\ &= \frac{-sin(x)}{sin(x) + cos(x)} = \frac{cos(x) sin(x) -cos^2(x)}{[sin(x) + cos(x)]^2} \\ \\ &= \frac{-1 - 2cos(x) sin(x)}{[sin(x) + cos(x)]^2} \end{align}$$

$$f(x) = x \cdot sin \left( \frac{1}{x}\right) = x \cdot sin(x^{-1})$$

$$f'(x) = (1)\cdot sin(x^{-1}) + x\cdot cos (x^{-1})(-x^{-2})$$

$$= sin \left( \frac{1}{x} \right) - \frac{1}{x^2} \cdot cos \left( \frac{1}{x} \right)$$

$$f(x) = sin(sin(sin(x)))$$

$$f'(x) = cos(sin(sin(x))) \cdot cos(sin(x)) \cdot cos(x)$$

$$f(\theta) = 2 cot(n\theta)$$

$$f'(x) = 2 n\cdot csc^2 (n \theta)$$

Here again, expressing the radical using a rational exponent makes this one easier:

$$f(x) = \sqrt{\frac{x}{x^2 + 4}} = \left( \frac{x}{x^2 + 4} \right)^{\frac{1}{2}}$$

$$\begin{align} f'(x) &= \frac{1}{2} \left( \frac{x}{x^2 + 4} \right)^{-\frac{1}{2}} \left( \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} \right) \\ \\ &= \frac{1}{2} \left( \frac{x^2 + 4}{x}) \right)^{\frac{1}{2}}\left( \frac{4 - x^2}{(x^2 + 4)^2} \right) \\ \\ \end{align}$$