So far we've seen how to *differentiate a function* that was made of one single term, like \(f(x)=3x^4\), for which \(f'(x)=12x^3\).

But what if we had to *differentiate a function that had 2, or 3, or more, terms*?

For instance, how would we differentuate the function: \[y = 3x^5+4x^2\] which consists of 2 terms being added?

Given two *functions* \(f(x)\) and \(g(x)\), to *differentiate* their *sum*, or their *difference*, we *differentiate* each of them
terms as though they were on their own:
\[\begin{pmatrix} f(x) \pm g(x) \end{pmatrix}' = f'(x) \pm g'(x) \]
In other words the *derivative of the sum, or the difference, of two functions is equal to the sum, or the difference, of the derivatives of each of the two functions*.

*Differentiate* each of the following *functions* with respect to \(x\):

- \(f(x)=4x^5+7x\)
- \(y=2x^3-2\sqrt{x}\)
- \(f(x)=\frac{3}{x} + x^4-7\)

Given a function \(f(x)\) and a scalar \(\alpha \in \mathbb{R}\) (a real number) the *derivative* of the *product* \(\alpha f(x)\) equals
to the *product* of \(\alpha \) and the *derivative* \(f'(x)\).

That's:
\[\begin{pmatrix} \alpha f(x) \end{pmatrix}' = \alpha f'(x)\]
Put simply: *when we differentiate a function that is being multiplied by a number, the result is the derivative of that function multiplied by that same number.*

Given two functions \(f(x)\) and \(g(x)\) as well as two scalars \(\alpha\) and \(\beta \) both in \(\mathbb{R}\), the derivative
of any linear combination of \(f(x)\) and \(g(x)\) equals to the same linear combination of their derivatives.

That's:
\[\begin{pmatrix} \alpha f(x) + \beta g(x) \end{pmatrix}' = \alpha f'(x) + \beta g'(x)\]

Given \(f(x)=x^3\), \(g(x)=\sqrt{x}\) and \(h(x) = \frac{1}{x^2}\), differentiate each of the following:

- \(3f(x)+4g(x)\)
- \(g(x)-6h(x)\)
- \(\frac{5}{6}f(x)+\frac{1}{8}h(x)-\frac{3}{4}g(x)\)