Understanding Differentiation with the Product Rule

10 Questions

What is the derivative of the function $u(x) = x^2 an x$?

$ an x - 2x an x$

If $f(x) = x an x$, what is $f'(x)$?

$ an x + x an^2 x$

Given $g(x) = x^3 ext{e}^x$, what is $g'(x)$?

$3x^2 ext{e}^x + x^3 ext{e}^x$

What is the derivative of $h(x) = x^4 ext{ln }x$?

$4x^3 ext{ln }x + 1/x$

If $k(x) = x^2 e^{2x}$, what is $k'(x)$?

$2xe^{2x} + 2x^2 e^{2x}$

What is the value of $x$ if $x^2 + 1 = 0$?

$x = i$

Simplify the expression: $(x + 1)^2 - (x^2 + 1)$

$2$

What does the expression $rac{d}{dx} [x(x-1) - x(x+1)]$ simplify to?

$-2$

What is the result of the expression: $(x+1)(2-1) + (x+1)(x-1)$?

$3x^2 - x - 1$

Simplify the expression: $rac{d}{dx} [x(x+1) - x^2]$

$1$

Study Notes

Understanding Differentiation through the Product Rule

Differential calculus is all about figuring out how functions change as you move along their graphs. One fundamental concept within this field is differentiation—the process of finding derivatives, which represent the slopes of curves at any given point. And while there's a straightforward formula for deriving simple functions like (x^n), things can become more complex when dealing with products of two or more functions. This complication calls for a specialized technique called the product rule.

Let's start by considering a function consisting of the product of two simpler functions, say (f(x) = g(x)\cdot h(x)). By definition, the derivative of such a composite function is calculated using the following equation:

[ f'(x) = \frac{d}{dx}(g(x)h(x)) = g'(x)h(x) + g(x)h'(x) ]

This is our product rule! It helps us find the derivative of a product of functions, even if we don't know the rule for the combined function itself.

Now let's break down each part of the product rule formula:

The first term: (g'(x)h(x)): This represents the derivative of one of the original functions multiplied by the other function evaluated at its current value.
The second term: (g(x)h'(x)): Here we have the initial function multiplied by the derivative of the second function.

By adding these two parts together, we obtain the total derivative of the composed function.

Here's a step-by-step example using sinusoidal functions: Let's calculate the derivative of (u(x)= x \sin x):

Firstly, note that (f(x) = u(x) = g(x)h(x) = x \cdot \sin x). Applying the product rule gives:

[ u'(x) = \frac{d}{dx} \left( x\sin x \right) = (\frac{d}{dx} x)(\sin x)+(x)\left( \frac{d}{dx}\sin x \right ) ]

Remembering the properties of differentials, we get:

[ u'(x) = (1)(\sin x) + (x)(-\cos x) ]

Simplifying further yields:

[ u'(x) = \sin x - x\cos x ]

So now we know the derivative of (x\sin x). In general, when applying the product rule, remember that:

Derive only one factor inside the parentheses, multiplying it with the entire expression outside the parentheses.
Multiply both factors in the other term.
Add up the results from steps 1 and 2.

While the product rule may seem intimidating at first glance, practice will help solidify your understanding of it. Remember, once you understand the basic rules of differentiation, including the product rule, its applications across various fields become apparent and vital skills develop naturally.