Derivative Rules
In calculus, a derivative can be thought of as an instantaneous rate of change; that is, how much a quantity is changing at a given point. Let’s take a closer look at how we can differentiate a function easily by the use of some helpful rules.
Understanding the Derivative
\(\)
Differentiation is a method to compute the rate at which a dependent variable y changes with respect to the change in the independent variable \(x\). This rate of change is called the derivative of \(y\) with respect to \(x\). There are many different notations to denote “take the derivative of.” The relationship between \(y\) and \(x\) is usually denoted by \(f(x)\) and its derivative is usually denoted as \( f'(x)\) or \(y’\) or \(\frac{dy}{dx}\). The definition of the derivative is given by:
\[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]
which is a lengthy procedure used to evaluate the derivative of a function. Thankfully, easier methods have been developed to help evaluate derivatives more quickly.
The Power Rule
If n is any real number, then \(\frac{d(x^n)}{dx}=nx^{n-1}\).
Also, remember that the derivatives of a constant is zero: \(\frac{d(c)}{dx}=0\).
Note that the sum and difference rule states:
\(\frac{d}{dx}[f(x)+g(x)-h(x)] = \frac{d}{dx}f(x)+\frac{d}{dx}g(x)-\frac{d}{dx}h(x)\)
(Just simply apply the power rule to each term in the function separately).
Example: Find the derivative of
\(f(x)=4-4\sqrt{x}+\frac{2}{x}\)
Solution: First, rewrite the function so that all variables of \(x\) have an exponent in the numerator:
\(f(x)=4-4x^{\frac{1}{2}} + 2x^{-1}\)
Now, apply the power rule to the function:
\(f'(x)=0-4\cdot\frac{1}{2}x^{\frac{-1}{2}}+2\cdot(-1)x^{-2}\)
Lastly, simplify your derivative:
\(f'(x)=-\frac{2}{\sqrt{x}} - \frac{2}{x^2}\)
The Product Rule
We already know how to find the derivative of a sum or difference of functions, but what about the product of two functions? The product of two functions is when two functions are being multiplied together. If \(f\) and \(g\) are both differentiable, then the product rule states:
\(\frac{d}{dx}[f(x)g(x)] = \frac{d}{dx}[f(x)]g(x) + f(x)\frac{d}{dx}[g(x)]\)
Example: Find the derivative of h(x) = (3x + 1)(8x4 +5x).
Solution: Using the above formula, let \(f(x) = (3x+1)\) and let \(g(x) = (8x^4 + 5x) \).
Now, find
\(\frac{d}{dx}[f(x)]\) and \( \frac{d}{dx}[g(x)]: \frac{d}{dx}[f(x)]=3\) and \( \frac{d}{dx}[g(x)] = 32x^3+5\)
Lastly, apply the product rule using the above formula:
\(h'(x) = \frac{d}{dx}[(3x+1)(8x^4+5x)] = \frac{d}{dx}[f(x)]g(x)+f(x)\frac{d}{dx}[g(x)]\)
\(h'(x) = (3)(8x^4+5x)+(3x+1)(32x^3+5)\)
The Quotient Rule
Now, let's take a look at the quotient of functions (when two functions are being divided by each other). If \(f\) and \(g\) are both differentiable, then the quotient rule states:
\(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\frac{d}{dx}[f(x)]g(x)-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}\)
Example: Find the derivative of
\(h(x) = \frac{3x-4}{2x^2-1}\)
Solution: Using the above formula, let \(f(x) = (3x-4)\) and let \(g(x) = (2x^2-1)\).
Now, find
\(\frac{d}{dx}[f(x)]\) and\(\frac{d}{dx}[g(x)]:\) \(\frac{d}{dx}[f(x)]=3\) and \(\frac{d}{dx}[g(x)]=4x\)
Lastly, apply the quotient rule using the above formula:
\(h'(x) = \frac{d}{dx}\left[\frac{3x-4}{2x^2-1}\right] = \frac{\frac{d}{dx}[f(x)]g(x)-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}\)
\(h'(x) = \frac{(3)(2x^2-1)-(3x-4)(4x)}{[(2x^2-1)]^2}\)
The Chain Rule
If \(f\) and \(g\) are both differentiable and \(F\) is the composite function defined by \(F(x) = f(g(x))\), then \(F\) is differentiable and the chain rule can be applied to \(F'\), which is given by \(F '(x) = f '(g(x))g '(x)\)
Example: Find the derivative of \(F(x) = (3x^4+2x^2-7)^5\).
Solution: Using the above formula, let \(g(x) = (3x^4 + 2x^2 - 7)\) and let \(f(x) = (x)^5\).
Now, find \(f '(x)\) and \(g'(x)\): \(f'(x) = 5x^4\) and \(g'(x) = 12x^3 + 4x - 0\)
So, if \(f'(x) = 5x^4\), then the composite function \(f'(g(x)) = 5(3x^4 + 2x^2 - 7)^4\)
Lastly, apply the chain rule using the above formula:
\(F'(x) = f'(g(x))g'(x)= 5(3x^4+2x^2-7)^4(12x^3+4x)\).
