The original motivation for studying derivatives came from physics, at a time when people were trying to understand things in motion. In this context, the derivative measures velocity, which we can think of as an instantaneous rate of change. The acceleration of an object measures how fast its velocity is changing, and so this can also be measured by a derivative (the second derivative).
When working with the graph of a function, the derivative can be very useful. For a straight line, the derivative is just the slope of the line, and a positive slope corresponds to a graph that moves up (from left to right), while a negative slope corresponds to a graph that moves down. For more general functions, just knowing the sign of the derivative is useful, since the graph goes up when the derivative is positive, and down when it is negative.
The very simple idea of measuring growth as either positive or negative makes it possible to find maximum and minimum values for a function. If the growth of a function changes from positive to negative, then you know that it has a (relative) maximum value when it changes. If the growth changes from negative to positive, then the function has a (relative) minimum value when it changes. So if we can find a function to describe a relationship in real life, we can predict where the maximum and minimum values will occur. Some of the applications will involve writing down a function to describe the profit for a company, and then deciding on the strategy that will maximize the profit.
The formula rate x time = distance should be familiar to you. This works only when the rate stays constant. Using calculus, we can deal with situations in which the rate is changing, since the concept of an integral allows us to find an average rate and use that to find the total distance traveled. Similarly, if we have a function that describes the rate of output of some process, we can find the total output even if the rate is variable, since we can calculate the average rate of output and multiply this average value by the total time. and multiply this average value by the total time.
There is a longer essay online, if you have the time to read it.
To take the derivative of the sum of two functions, we use the rule that the derivative of a sum is the sum of the derivatives. Unfortunately, the rule for taking the derivative of a product of two functions is more complicated. A simple example for which we already know the answer gives a good illustration of the difficulties.
Suppose that f(x) = x^{2} and g(x) = x^{3}. Then the individual derivatives are f '(x) = 2x and g '(x) = 3x^{2}. We know that the derivative of the product f(x)g(x) = x^{2}x^{3} = x^{5} should be 5x^{4}, but it is difficult to see how to combine f '(x) and g '(x) to get this answer. It turns out that we also have to use f(x) and g(x) in the formula.
One approach to understanding the formula for differentiating products comes from working with tangent lines. Remember that the equation of the line through the point (a,b) with slope m is
y = m(x - a) + b.
At the point (a,f(a)) on the curve y = f(x), we have the y-coordinate b = f(a), and the slope of the tangent line at this point is given by m = f '(a). This allows us to calculate the equation of the line tangent to the curve y = f(x) at x=a asy = f '(a)(x - a) + f(a).
In the same way, the line tangent to the curve y = g(x) at x=a isy = g '(a)(x - a) + g(a).
To get the formula for the tangent line to y = f(x)g(x) at x=a, we could try multiplying the two formulas. We would get the following quadratic expression.f '(a)g '(a)(x - a)^{2} + ( f '(a)g(a) + f(a)g '(a) )(x - a) + f(a)g(a)
Because the tangent line has to be linear, this doesn't solve the problem, unless we can change it in some way. The tangent line is intended to be the line that gives the best approximation to the curve, near the point (a,f(a)). When x is very close to a, the factor (x - a) is very small. In comparison, the factor (x - a)^{2} is even smaller; in fact, it is very much smaller. For example, if x - a = .1, then (x - a)^{2} = .01; if x - a = .01, then (x - a)^{2} = .0001; if x - a = .001, then (x - a)^{2} = .000001. Because the tangent line is just an approximation to the curve, it turns out that we can ignore the quadratic term. Notice that the part we had to ignore contains the term f '(a)g '(a), which would have given us a "nice" formula.Conclusion: the tangent line to y = f(x)g(x) at x=a is
y = ( f '(a)g(a) + f(a)g '(a) )(x - a) + f(a)g(a).
The general formula is best given using x instead of a.
d/dx ( f(x)g(x) ) = f '(x)g(x) + f(x)g '(x).
Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.
The derivative of the product of two functions is equal to the derivative of the first times the second, plus the first times the derivative of the second.
Example 1. Going back to the first example, we can now compute
d/dx ( x^{2} x^{3} ) = (2x)(x^{3}) + (x^{2})(3x^{2}) = 2x^{4} + 3x^{4} = 5x^{4}.
Our formula gives the expected answer, but has to include f(x) and g(x) along with f '(x) and g '(x).Example 2. It is interesting to see the formula for the derivative of the product of three functions. It clearly shows the pattern in which we add up terms where we differentiate one function at a time.
d/dx ( f(x)g(x)h(x) ) = f '(x)g(x)h(x) + f(x)g '(x)h(x) + f(x)g(x)h '(x)
The rule for taking the derivative of the composition of two functions is also complicated. Again, a simple example for which we already know the answer gives a good illustration of the difficulties.
Suppose that f(y) = y^{3} and u(x) = x^{4}. The formula for the composite function is f(u(x)) = (x^{4})^{3}. The individual derivatives are f '(y) = 3y^{2} and u '(x) = 4x^{3}. We know that the derivative of the composite function f(u(x)) = x^{12} should be 12x^{11}. In this case we just need to substitute y=u(x) into f '(y) and multiply by u '(x). We can show this by looking at the tangent lines.
The line tangent to the curve y = u(x) at x=a is
y = u '(a)(x - a) + u(a).
At the point (b,f(b)) on the curve z = f(y), the slope of the tangent line at this point is given by f '(b). Thus the equation of the line tangent to the curve z = f(y) at y=b asz = f '(b)(y - b) + f(b).
To compute the composite function f(u(x)), we substitute the formula for u(x) into the formula for f(y). If x=a, then we evaluate f(y) at y=u(a). This means that in the equation of the tangent line, we must take b=u(a). The equation becomesz = f '(u(a))(y - u(a)) + f(u(a)).
To get the formula for the tangent line to z = f(u(x)) at x=a, we can substitute the formulay = u '(a)(x - a) + u(a)
into the formulaz = f '(u(a))(y - u(a)) + f(u(a)).
This gives usz = f '(u(a))( u '(a)(x - a) + u(a) - u(a) ) + f(u(a))
orz = f '(u(a)) u '(a) (x - a) + f(u(a))
The derivative is the slope f '(u(a)) u '(a). The general formula is best given using x instead of a.
d/dx ( f(u(x)) ) = f '(u(x)) u '(x).
Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.
The derivative of the composite of two functions is found by taking the derivative of the first function and substituting in the second, the multiplying by the derivative of the second.
Example. 1 Going back to the first example, we can now compute
d/dx ( (x^{4})^{3} ) = 3(x^{4})^{2}(4x^{3}) = (3x^{8})(4x^{3}) = 12x^{11}.
Our formula gives the expected answer. It is important to remember to multiply the term f '(u(x)) by u '(x).Example. 2 To give another point of view, we can check the chain rule in a case where we can actually apply the product rule. If f(u(x)) = u(x)^{3}, we can write
f(u(x)) = u(x)^{3} = u(x)u(x)u(x),
and so from the product rule we getf '(u(x)) = u '(x)u(x)u(x) + u(x)u '(x)u(x) + u(x)u(x)u '(x).
Combining terms gives us the correct answerf '(u(x)) = 3 u(x)^{2} u '(x).
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