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Series 3 and 4 contain exercises that focus on differentiation techniques, i.e. on ways to find the derivative of a function.

Main tools are the derivation rules from block II learning sequence 2 of the script.

Some of these exercises are solved in detail here:

### Series 3 exercise 1 h)

$f(x) = - 2{x^4} + 3{x^3} - 0.5{x^2} - 0.5x + 4$

### Series 3 Exercise 1 m)

$\displaystyle{f(x) = \frac{{2x - 5}}{{x + 3}}}$

### Series 3 Exercise 5 b)

$h(t) = \sqrt {5t} + 2$

### Series 4 Exercise 4 g)

$f(x) = \ln (2{x^3} - 1)$

### Series 4 Exercise 5 e)

$I(C) = (C + 2) \cdot {e^C}$

### Structure of the problem

To find the derivative of a function you need to know the derivation rules (=differentiation rules) we studied in block II learning sequence 2 of the script. These rules are listed up here.

To put these rules to work, you need to recognize the algebraic structure of the terms a functional equation consists of. Is this a sum? Are there inner and outer functions and how can you tell between them?

In case this should give you trouble, we recommend you have a look at the preparatory course that helps you get acquainted with variables, terms, formulas, and equations. The Introduction section might be enough to help you.

### Series 3 Exercise 1 h)

$f(x) = - 2{x^4} + 3{x^3} - 0.5{x^2} - 0.5x + 4$

This is a polynomial function (sometimes just called polynomial). Its basic structure is a sum of terms. The sum rule tells us that we can differentiate each of these terms, then add up the results, the sum of which is the derivative of f.

The summand terms themselves are power functions. Here, the power rule tell us how to proceed: take the exponent in front of the power term, keep the power function itself, but with the exponent diminished by one. The result is itself a power function, and it is the first derivative we have been looking for.

Here is how to proceed:

$\begin{array}{left}f'(x) = - 2 \cdot ({x^4})' + 3 \cdot ({x^3})' - 0.5 \cdot ({x^2})' - 0.5 \cdot (x)' + (4)'\\ \; \qquad = - 2 \cdot 4{x^3} + 3 \cdot 3{x^2} - 0.5 \cdot 2x - 0.5 \cdot 1 + 0\\ \; \qquad = - 8{x^3} + 9{x^2} - x - 0.5\end{array}$

### Series 3 Exercise 1 m)

$\displaystyle{f(x) = \frac{{2x - 5}}{{x + 3}}}$

This is a rational function, i.e. a quotient of two polynomial functions, in this case linear functions. The quotient rule shows how to find the derivative of this function. We apply this rule:

$\displaystyle{ f'(x) = \frac{{(2x - 5)' \cdot (x + 3) - (2x - 5) \cdot (x + 3)'}}{{{{(x + 3)}^2}}} }$

$\displaystyle{\qquad \; = \frac{{2 \cdot (x + 3) - (2x - 5) \cdot 1}}{{{{(x + 3)}^2}}}}$

In principle, we can leave our result like that. It might, however, be possible to simplify terms. It would make little sense to expand the denominator. By contrast, expanding the numerator helps us eliminate all the terms containing x:

$\displaystyle{f'(x) = \frac{{2x + 6 - 2x \,{\color {red} +} \,5}}{{{{(x + 3)}^2}}} = \frac{{11}}{{{{(x + 3)}^2}}}}$

### Series 3 Exercise 5 b)

$h(t) = \sqrt {5t} + 2$

Here we have a sum, the first term of which being a root function (see Derivatives of Power Functions), and the second a constant (which disappears when differentiating).

To differentiate the first term, we have two options:

Option 1:

We take the term $\sqrt {5t}$ as a composition (i.e. chain of functions, or linked functions), with the root function as an outer function, and $t \rightarrow 5t$ as an inner function. The chain rule allows us to differentiate the composition.

To make things easier, we rewrite the root as a power with exponent 1/2 :

$\displaystyle{h(t) = {(5t)^{^{\tfrac{1}{2}}}} + 2}$

Now we can differentiate the outer function according to the power rule, and the inner function according to the constant factor rule (inner function in blue):

$\displaystyle{h'(t) = \tfrac{1}{2} \cdot {(5t)^{^{ - \tfrac{1}{2}}}} \color {blue}{\cdot 5} + 0 = 2.5 \cdot {(5t)^{^{ - \tfrac{1}{2}}}}}$

Option 2:

We rewrite the entire first term so as to obtain a product of two root terms:

$\sqrt {5t} = \sqrt {5} \cdot \sqrt {t}$

The first term is a constant, while the second is a root function which we know how to differentiate (here again, apply the constant factor rule, but this time to the square root of 5):

$\displaystyle{h'(t) = \sqrt {5} \cdot \frac{1}{{2 \cdot \sqrt {t }}} + 0 = \frac{{\sqrt {5 }}}{{2 \cdot \sqrt {t} }}}$

A little extra exercise: show that the results obtained by both options are equivalent.

### Series 4 Exercise 4 g)

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

This is a typical composite or linked function:

First, x is transformed via an inner function into an intermediate result:

$u=2{x^3} - 1$

Then, this intermediate result will be transformed via an outer function into yet another term. This outer function is the natural logarithm.

The chain rule says that the derivative of the composite function is the derivative of the outer function (blue) times the derivative of the inner function (red).
Watch the arguments to avoid errors!

$\displaystyle{f'(x) = \color {blue}{\frac{1}{{2{x^3} - 1}}}\cdot \color {red} {6{x^2}} = \frac{{6{x^2}}}{{2{x^3} - 1}}}$

### Series 4 Exercise 5 e)

$\displaystyle{I(C) = (C + 2) \cdot {e^C}}$

Here we have the product of a linear function (terms in parentheses) and an exponential function. The product rules says that we need each of these functions, and their derivatives, and put them together in a certain way:

$I'(C) = 1 \cdot {e^C} + (C + 2) \cdot {e^C}$

We might leave it like that, but for certain purposes, like when setting the first derivative equal to zero for finding an extremum, it makes sense to factor out.

This can be done to the term ${e^C}$, leading to:

$\displaystyle{I'(C) = 1 \cdot {e^C} + (C + 2) \cdot {e^C} = (1 + (C + 2)) \cdot {e^C} = (C + 3) \cdot {e^C}}$

### Series 3 Exercise 1 h)

$f(x) = - 2{x^4} + 3{x^3} - 0.5{x^2} - 0.5x + 4$

$f'(x) = - 8{x^3} + 9{x^2} - x - 0.5$

### Series 3 Exercise 1 m)

$\displaystyle{f(x) = \frac{{2x - 5}}{{x + 3}}}$

Quotient rule:

$\displaystyle{f'(x) = \frac{{2 \cdot (x + 3) - (2x - 5) \cdot 1}}{{{{(x + 3)}^2}}} = \frac{{2x + 6 - 2x + 5}}{{{{(x + 3)}^2}}} = \frac{{11}}{{{{(x + 3)}^2}}}}$

### Series 3 Exercise 5 b)

$h(t) = \sqrt {5t} + 2$

$\displaystyle{h(t) = {(5t)^{^{\tfrac{1}{2}}}} + 2}$

$\displaystyle{h'(t) = \tfrac{1}{2} \cdot {(5t)^{^{ - \tfrac{1}{2}}}} \cdot 5 = \frac{5}{{2 \cdot \sqrt {5t} }} = \frac{{\sqrt {5 }}}{{2 \cdot \sqrt {t} }}}$

### Series 4 Exercise 4 g)

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

Composite function, thus chain rule:

$\displaystyle{f'(x) = \frac{1}{{2{x^3} - 1}}\cdot 6{x^2}} = \frac{{6{x^2}}}{{2{x^3} - 1}}$

### Series 4 Exercise 5 e)

$\displaystyle{I(C) = (C + 2) \cdot {e^C}}$

With product rule:

$\displaystyle{I'(C) = 1 \cdot {e^C} + (C + 2) \cdot {e^C} = (1 + (C + 2)) \cdot {e^C} = (C + 3) \cdot {e^C}}$