Differentiation Problems

Finding derivatives from the formula is possible but cumbersome, so we'll look at a few rules that make life a lot easier.

Using the Definition

The basic way to find derivatives is to use its definition:

\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)} {h} \]

so the derivative of \(f(x) = \frac{1}{x}\) is:

\[ \begin{align} f'(x) &= \lim_{h \to 0} \frac {\frac{1}{x + h} - \frac{1}{x}} {h} \\ &= \lim_{h \to 0} \frac{\frac{x - (x + h)}{x(x + h)}} {h} \\ &= \lim_{h \to 0} \frac{-h} {hx(x + h)} \\ &= \lim_{h \to 0} \frac{-1} {x(x + h)} \\ &= -\frac{1}{x^2} \end{align} \]

In fact, if we replace \(x^n\) to \(x\) to get \(f(x) = x^n\) we can get more general equation:

\[ \begin{align} f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)} {h} \\ &= \lim_{h \to 0} \frac{(x + h)^n - x^n} {h} \\ &= \lim_{h \to 0} \frac{x^n + nhx^{n - 1} + h^2 \times (something...) - x^n} {h} \\ &= \lim_{h \to 0} \frac{nhx^{n - 1} + h^2 \times (something...)} {h} \\ &= \lim_{h \to 0} (nx^{n - 1} + h \times (something...)) \\ &= nx^{n - 1} \end{align} \]

Using rules

Let's define a function \(f\) as following:

\[ f(x) = \frac{3x^7 + x^4\sqrt{2x^5 + 14x^{\frac{4}{3}} - 23x + 9}} {6x^2 - 4} \]

The key to differentiating a function is to understand how it is synthesized from simpler functions. Here are some rules we can use.

Constant multiples of functions

You just multiply by the constant after your differentiate.

\[ (nf(x))' = nf'(x) \]

Sums and differences of functions

You just differentiate each piece and then add or subtract.

\[ (f(x) + g(x))' = f'(x) + g'(x) \]

Products of functions via product rule

To solve the problems involving products of functions, you have to mix and match. That is the product rule(version 1):

\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \]

If we replace \(f(x)\) with \(u\) and \(g(x)\) with \(v\), we got(version 2);

\[ \frac{dy} {dx} = v\frac{du}{dx} + u\frac{dv}{dx} \]

What if you have a product of three terms? We have the product rule for three terms(three variable):

\[ \frac{duvw} {dx} = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx} \]

Quotients of functions via the quotient rule

Quotients are handled in a way similar to products, except that the rule is a little different. The form of quotients is \(h(x) = \frac{f(x)}{g(x)}\), here is the quotient rule(version 1):

\[ (\frac{f(x)} {g(x)})' = \frac{f'(x)g(x) - f(x)g'(x)} {(g(x))^2} \]

There is also another version(version 2):

\[ \frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{x} - u\frac{dv}{dx}} {v^2} \]

Composition of functions via the chain rule

The Chain rule is discrebed as(version 1):

\[ (f(g(x)))' = f'(g(x))g'(x) \]

And replace \(f\) and \(g\) with \(y\) and \(u\), we can invoke the other version of the chain rule(version 2):

\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} \]

Derivatives of Piecewise-Defined Functions

Consider the following piecewise-defined function:

\[ f(x) = \begin{cases} 1, &if \quad x \le 0, \\ x^2 + 1, &if \quad x > 0. \\ \end{cases} \]

To check that a piecewise-defined function is differentiable at a point where the pieces join together, you need to check:

• pieces agree at the join point(for continuity);
• the derivatives of the pieces agree at the join point.

Otherwise it's not differentiable at the join point.