d/dx(sin(x))

[danbaron]

As I remember, in calculus there are three main concepts, limits, derivatives, and integrals.

In words, the definition of the derivative (d/dx) of the function, f(x), is the slope of the tangent line to f(x), at x.

("d/dx" is a symbol, meaning, "the derivative with respect to x". It doesn't mean, "d divided by dx". All of the characters taken together of "d/dx", constitute one symbol.)

So, for instance,

"d/dx(cos(x))", means, "the derivative of the function cos(x), with respect to x".

The tangent line to a curve at a point p is defined as a (straight) line which touches the curve at p, but does not cross the curve.

For a straight line, the line and its tangent are coincident.

For instance, for the line, y = 2x, its slope is 2, meaning its derivative is 2 at any point on the line.

For any line, the slope, and therefore the derivative are found by taking any two points on the line, p0 and p1.

slope, = derivative with respect to x, = (y1-y0)/(x1-x0).

Similarly, the tangent of the acute angle t (tan(t)) of a right triangle is, sin(t)/cos(t).

The slope of a line is often referred to as, "the rise over the run".

The general form of any straight line is,

y = m*x+b.

"m" is the slope (derivative), and "b" is the y intercept (the value of y when x = 0).

So, for straight lines, finding the derivative is trivial.

But, how about finding derivatives of functions which are curves?

Symbolically, the definition of the derivative is,

d/dx(f(x)) = the limit as e approaches 0 of, (f(x+e)-f(x))/e.

If you look at the picture (attached file, "derivative.jpg"), say we want to find the tangent line to the curve at the point, x0.

To do that, we define another point at x1.

The x distance between x0 and x1, we call e.

(f(x1) corresponds to f(x+e), and, f(x0) corresponds to f(x).)

We keep x0 fixed, and we decrease e, moving x1 closer and closer to x0.

As we do this, the approximate tangent line becomes closer and closer to, the actual tangent line to the curve at x0.

In the limit as e approaches 0, x0 and x1 become the same point, and we then have found the slope of the curve at x0, and therefore its derivative at x0.

We can do an example and show that when we symbolically determine the derivative of a function, f(x), it applies to every point on the curve.

Let's find the derivative of the function, y = x^2.

To do so, we plug it into the symbolic definition.

d/dx(x^2) = limit e-->0 ((x+e)^2-x^2)/e.

("e-->0" means, "as e approaches 0")

= limit e-->0 (x^2 + e^2 + 2*x*e - x^2)/e,

= limit e-->0 e^2/e + 2*x*e/e,

= limit e-->0 e + 2*x,

= 0 + 2*x,

= 2*x.

Therefore,

d/dx(x^2) = 2*x.

In words, the derivative of the x^2 with respect to x, is 2*x.

This means that for any point on the curve y = x^2, the slope at the point corresponding to x = x0, is 2*x0.

So, at the point (2,4), the slope of the curve is 4, and, at the point (3,9) the slope of the curve is 6, etc.

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Now let's try to find the derivative of a function which is not so easy, y = sin(x).

We try plugging it into the definition.

d/dx(sin(x)) = limit e-->0 (sin(x+e) - sin(x))/e.

As e-->0, the numerator approaches, sin(x) - sin(x), = 0,

and the denominator also approaches 0,

and we get, 0/0.

Therefore, the calculation is wrong.

Anytime you do a calculation and get 0/0, or, infinity/infinity, the answer is wrong.

So, we have to do some trigonometry before letting e-->0.

A trigonometric identity is,

sin(A+B) = sin(A)*cos(B) + cos(a)*sin(B).

Now we can proceed.

d/dx(sin(x)) = limit e-->0 (sin(x+e) - sin(x))/e.

= limit e-->0 (sin(x)*cos(e) + cos(x)*sin(e) - sin(x))/e,

= limit e-->0 (sin(x)*(cos(e) - 1) + cos(x)*sin(e))/e,

= limit e-->0 (-sin(x)*(1 - cos(e)) + cos(x)*sin(e))/e,

= limit e-->0 -sin(x)*(1 - cos(e))/e + cos(x)*sin(e)/e.

Now, if you look at the link,

http://www.thinbasic.com/community/showthread.php?11264-The-limit-as-t-gt-0-of-sin(t)-t

and there you change the variable "t" to "e", you see that,

limit e-->0 (1 - cos(e))/e = 0,

and,

limit e-->0 sin(e)/e = 1.

Therefore,

d/dx(sin(x)) = -sin(x)*0 + cos(x)*1, = cos(x).

So,

d/dx(sin(x)) = cos(x).

In words, the derivative of the sine is the cosine.

We can check a little bit to see if our answer makes sense.

When we plot sin(x), we see that at x = pi/2, the slope of the curve is 0.

It means that at x = pi/2, the value of the cosine should be 0, and fortunately, it is.

[kryton9]

Thanks Dan. I took Calculus and how I got through the class I have no memory of. Until Calculus I was an A- / B+ student in Math, but Calculus, I could never really grasp. I will refer to this post many times I am sure. I think with computers and being able to plot things it will be a lot easier to grasp.