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
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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 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 \(\frac{dx^n}{dx}=nx^{n-1}\).
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 formule, let f(x) = (3x-4) and let g(x) = (2x2-1).
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)\).
