Solution using the meaning of composed functions, the chain rule and the given data: 4. The hot glowing gas of a star radiates energy according to the function E = s T 4.If T is a function of the radius of the star, use the Chain Rule to find a formula for dE/dr. Solution, using the chain rule.
Online Derivative Calculator Solve derivatives with Wolfram Alpha Example input More than just an online derivative solverWolfram Alpha is a great calculator for first, second and third derivatives; derivatives at a point; and partial derivatives. Learn what derivatives are and how Wolfram Alpha calculates them.Learn more about:.Tips for entering queriesEnter your queries using plain English. To avoid ambiguous queries, make sure to use parentheses where necessary. Here are some examples illustrating how to ask for a derivative.Access instant learning toolsGet immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator.Learn more about:.VIEW ALL CALCULATORS.What are derivatives? The derivative is an important tool in calculus that represents an infinitesimal change in a function with respect to one of its variables.Given a function, there are many ways to denote the derivative of with respect to. The most common ways are. When a derivative is taken times, the notation or is used.
These are called higher-order derivatives. Note for second-order derivatives, the notation is often used.At a point, the derivative is defined to be. This limit is not guaranteed to exist, but if it does, is said to be differentiable at. Geometrically speaking, is the slope of the tangent line of at.As an example, if, then and then we can compute:. The derivative is a powerful tool with many applications. For example, it is used to find local/global extrema, find inflection points, solve optimization problems and describe the motion of objects.
How Wolfram Alpha calculates derivativesWolfram Alpha calls Wolfram Languages's function, which uses a table of identities much larger than one would find in a standard calculus textbook. It uses well-known rules such as the linearity of the derivative, product rule, power rule, chain rule and so on.
Additionally, uses lesser-known rules to calculate the derivative of a wide array of special functions. For higher-order derivatives, certain rules, like the general Leibniz product rule, can speed up calculations.
What I hope to do in this video is a proof of the famous and useful and somewhat elegant andsometimes infamous chain rule. And, if you've beenfollowing some of the videos on 'differentiability implies continuity', and what happens to a continuous function as our change in x, if x isour independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actuallysurprisingly straightforward, so let's just get to it, and this is just one of many proofs of the chain rule. So the chain rule tells us that if y is a function of u, which is a function of x, and we want to figure outthe derivative of this, so we want to differentiatethis with respect to x, so we're gonna differentiatethis with respect to x, we could write this as the derivative of y with respect to x, which is going to beequal to the derivative of y with respect to u, times the derivativeof u with respect to x. This is what the chain rule tells us. But how do we actuallygo about proving it? Well we just have to remind ourselves that the derivative ofy with respect to x. The derivative of y with respect to x, is equal to the limit asdelta x approaches zero of change in y over change in x.
Now we can do a little bit ofalgebraic manipulation here to introduce a changein u, so let's do that. So this is going to be the same thing as the limit as delta x approaches zero, and I'm gonna rewritethis part right over here. I'm gonna essentially divide and multiply by a change in u. So I could rewrite this as delta y over delta u times delta u, whoops. Times delta u over delta x.
Change in y over change in u, times change in u over change in x. And you can see, these arejust going to be numbers here, so our change in u, thiswould cancel with that, and you'd be left withchange in y over change x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to?
Well the limit of the product is the same thing as theproduct of the limit, so this is going to be the same thing as the limit as delta x approaches zero of,and I'll color-coat it, of this stuff, of delta y over delta u, times- maybe I'll put parentheses around it, times the limit. The limit as delta x approaches zero, delta x approaches zero, of this business. So let me put some parentheses around it.
Delta u over delta x. So what does this simplify to? Well this right over here,this is the definition, and if we're assuming, inorder for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in orderfor this to be true, we're assuming. We're assuming y commau are differentiable. Are differentiable at x.
![Rule Rule](http://3.bp.blogspot.com/-IFIZ52eBCnI/Va5IrIsuIZI/AAAAAAAAA24/-zgVPXZLQq4/s320/3.png)
And remember also, ifthey're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, andthis is the derivative of.
This is u prime of x, or du/dx, so this right over here. We can rewrite as du/dx, I think you see where this is going. Now this right over here, just looking at it the wayit's written out right here, we can't quite yet call this dy/du, because this is the limitas delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from, probably,the previous video depending on how you're watching it, which is, if we have a function u that is continuous at a point, that, as delta x approaches zero, delta u approaches zero. So we can actually rewrite this.
We can rewrite this right over here, instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta uis going to approach zero. As our change in x gets smallerand smaller and smaller, our change in u is going to get smaller and smaller and smaller. So we can rewrite this, as our change in u approaches zero, and when we rewrite it like that, well then this is just dy/du. This is just dy, the derivativeof y, with respect to u. So just like that, if we assume y and u are differentiable at x, or you could say thaty is a function of u, which is a function of x, we've just shown, infairly simple algebra here, and using some assumptions about differentiability and continuity, that it is indeed the case that the derivative of y with respect to x is equal to the derivativeof y with respect to u times the derivativeof u with respect to x.
Hopefully you find that convincing.