Differentiation and Integration of Sinusoidial functions
Here is a process to generating the derivatives (and subsequently, the anti-derivatives) of some basic trigonometric functions.
Differentiate is the operation which provides the gradient of a function. If once upon differentiation we obtain the derivative, the original function is referred to as the anti-derivative. Put another way, the anti-derivative of a function is its integral, and is obtained not through differentiation, but through integration. As we shall see, for basic trigonometric function, integration and differentation are closely related.
To begin with, we shall only consider the function of the form `cos(x)`, that is, without any additional arguments.
Obtaining the Derivative
If we want to work out the derivative of the cosine function, we can start by drawing out a cosine curve. We then draw on some tangents: those at the turning points (at the top of the crests and the bottom of the troughs), and then also at the steepest parts of the curve, which occur when the curve cuts the x-axis.
Figure: `cos(x)` with some tangents
Assigning values for these tangents, `m`: 0 for the peaks and troughs, +1 when the wave is on the incline and -1 when the wave in on the decline. Using these values, draw another graph on which we plot those tangents [-1, 0, +1]. Then we sketch a curve which goes through each of the points just plotted, and we obtain:
Figure: the derivative of `cos(x)`, `y=-sin(x)`
What we have just achieved is finding the derivative. Notice that we have effectively just shifted the curve, one quarter of its cycle, to the left. We should note, that the curve we 've just drawn is actually `y=-sin(x)`. We can again repeat the process, first drawing on tangents and assigning them a magnitude,
Figure: -sin(x) with some tangents
...again plotting those tangents on a new graph and sketching a curve through them
Figure: The derivative of `-sin(x)`, `y=-cos(x)`
Obtaining the Integral
As mentioned, the finding the integral of a trigonometric function can be thought of (loosely) as the opposite operation to differentation. As we've seen, differentiation of a sinusoid shifts the curve to the left - hence, integration, intuitively, shifts the function to the right. If differentation shifts the sinusoid to the left, integration shifts it to the right.
The process for more complex arguments
In science it is rare to find a trigonometric function with only the dependent variable as an argument. In most situations we are presented with a problem involving additional arguments, such as `y=cos(kx)` or `y=cos(kx+c)`. To go about differentiating these functions, we employ the method of u-substitution. This is an expression of the chain rule.
To integrate and differentiate these such functions, we first make the expression conform to a more simple form. For example,
`(dcos(kx+c))/(dx)=?`
First we make the substitution, `u=kx+c`. This makes the cosine function become, `cos(kx+c)=cos(u)`. So now we have the expression,
`(dcos(u))/(dx)=?`
The next part involves expressing `dx` in terms of `u`. To do this, we differentiate `u`,
`(du)/(dx)=(d(kx+c))/(dx)`
Differentiate is a linear operation, which means we can expand the brackets. Where c is a constant, its derivative is zero (it does not change with x), ie.,
`(du)/(dx)=(dkx)/(dx)+(dc)/(dx)=(dkx)/(dx)`
We now apply the product rule, which states that the derivative of the product of two functions is,
`(du)/(dx)=(dkx)/(dx)=k(dx)/(dx) + x(dk)/(dx)`
If k is a constant, then, again, its derivative is zero. Hence,
`(du)/(dx)=(dkx)/(dx)=k(dx)/(dx)`
Now cancel the `dx` terms, and obtain,
`(du)/(dx)=(dkx)/(dx)=k`
We rearrange the equation to make `dx` the subject, thus,
`dx=(du)/k`
We now have all of the essentials in place to return to the original problem,
`(dcos(kx+c))/(dx)=k(dcos(u))/(du)`
Having expressed the cosine in terms of only u, we can now use the differentiation cycle to find the derivative, ie., `cos(u) stackrel "differentiate" rarr -sin(u)`. Having performed the differentiation, we then reverse the substitution for u, thus,
`(dcos(kx+c))/(dx)=k(dcos(u))/(du)=-ksin(u)=-ksin(kx+c)`