Me parece que este video responde adecuadamente a la pregunta que nos hacemos :
Les comparto el acceso a un buen libro de cálculo:
¿Qué es el cálculo? VEAN ESTE VIDEO. LO RECOMIENDO:
Notas sobre derivación ( Tecnológico de Costa Rica )
Sean a, b y k constantes (números reales) y consideremos a: u y v como funciones.
En unos días vamos a poner ejercicios de derivadas implícitas, pero antes, les dejamos por aquí la definición.
En una correspondencia o también una función si está definida en forma implícita cuando no aparece despejada la y sino que la relación entre x e y viene dada por una ecuación la cual tiene de dos incógnitas cuyo segundo miembro es el cero.
Para poder hallar la derivada correcta en forma implícita no es necesario despejar y. Así que basta el derivar miembro a miembro paso por paso, utilizando así todas las reglas vistas hasta ahora en derivadas.es y teniendo presente lo siguiente:
En general y’≠1.
Por lo cual omitiremos x’ y dejaremos y’.
Y luego cuando las funciones son ya más complejas podemos utilizar una regla para facilitar el cálculo de la función:
1) ¿Qué es la diferenciación ?
2) Explica la Regla de la cadena en términos sencillos
Date: 10/12/96 at 18:6:19 From: Dominic Tsang Subject: Meaning of Derivative Can you give me a plain English meaning of the idea of a derivative? Thanks!
Date: 10/13/96 at 15:54:4 From: Doctor Scott Subject: Re: Meaning of Derivative Dominic, The derivative does, in fact, have a nice "plain English" meaning. You probably learned that the derivative is the slope of the tangent line to the curve at a point. The derivative describes the rate of change of the function at that point. Actually, it is the *instantaneous* rate of change of the function at the point. For example, if you are blowing up a balloon, the volume of the balloon depends on the radius of the balloon. That is, V = (4/3)*pi*r^3. The derivative of V (with respect to r) would tell you how fast the volume is changing as the radius changes. Hope this helps! -Doctor Scott, The Math Forum Check out our web site!
Date: 10/13/96 at 16:2:48 From: Doctor Ken Subject: Re: Derivative Hi Dominic - Here's an example of the derivative and what it means. Let's say you've got a function, call it f(x). The derivative of f, f'(x), tells you how fast f is changing. If the derivative is positive, f is increasing, and if the derivative is negative, f is decreasing. If the derivative is 0, then at that point f is neither increasing nor decreasing. Let's say f(x) represents the position of a car on a straight road at time x. Then the derivative f'(x) tells you the velocity (that's like speed) of the car at time x. If the car is going forward the velocity will be positive, and if it's going backward the velocity will be negative. If you take another derivative, you'll call it f''(x), and that tells you the acceleration of the car at time x. If the car is speeding up the acceleration will be positive, and if it's slowing down it will be negative. Here's a little-known fact I'll let you in on: if you take another derivative, that has a name too. It's called "jerk," and we write f'''(x). Can you figure out what the physical interpretation of jerk is? The derivative is used all the time in physics, in exactly this kind of way. -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 12 Jan 1995 12:57:41 -0500 From: Jenny Lay Subject: Help! Dear Dr. Math, Earlier this semester we learned about derivatives in Calculus. I know how to determine the derivative of something, but what is the purpose? Thanks. Jenny
Date: 12 Jan 1995 14:11:43 -0500 From: Dr. Ken Subject: Re: Help! Hello there! There are lots of reasons we'd want to take the derivative of something. First of all, let's say you're riding in your shiny new sports car and you have the best odometer in the world. It will tell you to the nearest thousandth of a mile (or something like that) how far you've gone. If you graphed what the odometer tells you as a function of time, so that time is on the x-axis and distance is on the y-axis, you could take the derivative of this function and figure out your speed for every point in your journey. So all the information about your speed and acceleration and everything can be gotten from the odometer, as long as you know how to take derivatives. Here's a question my calculus teacher once asked me: in cars, there's both an odometer and a speedometer. Essentially, the speedometer takes the derivative of the odometer information (before it gets to the odometer though; it's straight from the wheels). How does it do that? It's been doing that since way before on-board computers happened to cars. So essentially, they've found a purely mechanical way to take derivatives. Neat stuff, worth researching. The derivative is also quite an intuitive concept, I think. Let's say you have a growth chart on your wall. If you're a human (which I believe you are) you'll probably have a couple of periods when you grew faster than at other times in your life. If the marks were made at regular intervals, they'd be more spread out in certain periods and more clustered together in others. So it's not hard to figure out from this chart that you grew faster in those growth spurt times than in the lull times. Well, how fast you grew is just the derivative with respect to time of how tall you were. So the derivative will be big sometimes, small sometimes, and once you hit 40 years old, it will be negative (some people say). So these are a couple of real-life examples. Other examples that are based on integration (the inverse of differentiation) would include finding the volume of some objects, finding the area of some regions in a plane, and stuff like that. And trust me, if you go on and do some more in math, taking the derivative of functions will be SHEER BLISS compared with some of the more nasty stuff (which is more rewarding. Stick with math!). So that's how I feel about derivatives.
Date: 12 Jan 1995 15:23:35 -0500 From: Dr. Elizabeth Subject: Re: Help! Hi Jenny! One of the nicest things you can do with derivatives is to find out where the maximum value of a function is - which can be a very useful thing to know. The derivative of a function is its slope at any given point (actually it's the slope of the tangent line to the point, but even though a point can't have a slope, I always found it easier just to think of the derivative this way). At the highest point of a function (its maximum value), it's gone as far up as it's going to, and it's about to start heading back down. Its slope at this point, then, will be zero, since it isn't going up, and it isn't going down. Let's say you had a function and you wanted to know where, exactly, it would reach a maximum (I'll give you an example of why you might want to know this in a minute). Take the derivative of the function and set it equal to zero. Then solve for x or a or the number of feet of fence or the number of frogs or whatever it is that your independant variable happens to be. Here's my example: I throw a ball straight up at a speed of 6 meters per second somewhere where there's absolutely no air (no air resistance). When will it be at its highest point? I can write the ball's height in an equation: the height at any time t will be the ball's upward velocity times the amount of time it's been going up (6m/sec times time, t) minus its gravitational acceleration downward times the amount of time it's been up there squared (10 m/sec/sec *that's the acceleration of gravity* times time squared, t^2). Leaving out the units so that the math is easier to see here: Height = 6t-10t^2. If we take the derivative of this function, we have an equation for the slope of this function at any given time, t. The value of t for which this new equation is equal to zero is the same t at which the height of the ball will be a maximum. Derivative of height = 6-20t = 0 6 = 20t t = 6/20 second The ball will reach its maximum height in 6/20 of a second. Hope this helps! Elizabeth, a math doctor
Date: 11/23/97 at 20:09:15 From: stuart Subject: Calculus chain rule I can't understand the chain rule. Every time I ask someone to explain it they use y's and u's, etc... could you give me the chain rule in easy terms, like how to do it, not just give me a formula like y=(U)^2? Thanks. Stu
Date: 11/23/97 at 21:24:09 From: Doctor Scott Subject: Re: Calculus chain rule Hi Stu! Good question. I was just skimming an excellent Calculus book written by Paul Foerster where this very question was addressed. His suggestion was that you should think of the chain rule as a process rather than a rule with a lot of du/dx and dy/dx's. So, here goes.... Remember that the chain rule is used to find the derivative of *compositions of functions* - that is, functions that have functions inside of them. For example, the function sin(x^2) can be thought of as a composition of two other functions, sin x and x^2, with the x^2 being INSIDE the sin function. Similarly, the function (x^2 - 5x + 8)^(1/2) is also a composition of two other functions, (x^2 - 5x + 8) and x^(1/2), with the first function being INSIDE the second. One more example? The function cos(tan(5x-3)) is the composition of three functions, 5x - 3 inside of tan x, inside of cos x. So the chain rule gets applied when there is some function INSIDE of another function. The stuff that people have been telling you probably goes something like this: If y = sin(4x-3), then we can write this function as the composition of y = sin u and u = 4x - 3. (Again, notice that the 4x - 3 is INSIDE of the sin function.) Then, dy/dx = dy/du * du/dx. So, we have dy/dx = cos u * 4; but u = 4x-3, so we have dy/dx = 4cos(4x-3). How about another way? Let's think of the chain rule as a process. The derivative of a composite function is the DERIVATIVE OF THE OUTSIDE FUNCTION TIMES the DERIVATIVE OF THE INSIDE FUNCTION. In practice, here's how it works. Consider y = sin(4x-3). The outside function is a sine function; its derivative is cosine, so we have (so far) cos(4x-3). Now, INSIDE the sine function is 4x-3. Its derivative is 4, so now we have 4cos(4x-3). Notice that there is no other function "inside" the 4x-3, so we are done. Let's look at a couple more examples: y = (x^2 - 5x + 8)^(1/2). The OUTSIDE FUNCTION is basically a power rule problem, so we have 0.5(x^2 - 5x + 8)^(-1/2) using the power rule. The INSIDE FUNCTION is x^2 - 5x + 8; its derivative is 2x - 5, so we have y' = (2x - 5)(.5)(x^2 - 5x + 8)^(-1/2). y = cos(tan(5x-3)). The outermost function is a cosine, so its derivative is negative sine: -sin(tan(5x-3)). Inside the cosine is a tan function; its derivative is sec^2, so we now have sec^2 (5x-3) * (-sin(tan(5x-3)) Finally, inside of the tan function is 5x-3; its derivative is 5. So, FINALLY, we have 5 * sec^2 (5x-3) * (-sin(tan(5x-3)) Or, simplifying, we get y' = -5 sec^2 (5x-3) sin(tan(5x-3)) So, it helps a lot to think of the chain rule as: The derivative of the outside TIMES the derivative of what's inside! -Doctor Scott, The Math Forum Check out our web site! http://mathforum.org/dr.math/