tomta news, fearless in revealing the truth. zero product property definition, zero product property calculator, how to get free robux for free.
 Home
 About Us
 Contact Us
 Privacy policy
 Terms and conditions

Settings
 Dark mode
Derivative of secx tanx, The Ultimate Guide And What is the derivative of Secx
Labib
0 Commenter
What is the derivative of Secx
To get the derivative of the second,
What is a derivative of Secx
derivative of secx tanx. we can use either the limitation definition of the derivative (which can take a very long time) or the definition of the second itself:
Sec = 1 / cox
We know d / dx cosx
=  sinx  remember this because we will need it.
Our problem is: D / Dex Sex
Since sex = 1 / cox, we can take it as:
D / DX, 1 / Cox
Using the quotient rule we can get this derivative:
d / dx u / v = u'v  uv '/ v (square) 2
In our case, u = 1 → u '= 0 and v = cosx → v' = in synex
d / dx, 1 / coxx = (0) (coxx)  (1) ( synex) / (coxx) 2
= Synex / Cox 2x
This is equal to:
1
/ coxx ⋅ synex / coxx
1
/ cosx = secx and sinx / cosx = tanx because it is: sex, tanx
Therefore, d / dx secx = d / dx, 1 / cosx = secx, tanx
Derivative of secx tanx
we're going to talk about the secant method, we're going to derive the secant method of finding roots of nonlinear equations. derivative of secx tanx
So we're going to look at the derivation of how secant method is used to solve a nonlinear equation of the form f of x equal to 0.derivative of secx tanx
So if you have an equation where the lefthand side is f of x, so we're going to, let's suppose if we plot f of x, the lefthand side of the equation, which is the function, and we plot it as a function of x, and we find out where does it become 0, that becomes the root of the equation.
Also read
So that's what we want to look for, we want to see that where does f of x become 0, and that's what we're going to derive by using this numerical method called the secant method.derivative of secx tanx
Now, the approach which we're going to take in order to derive it is based on NewtonRaphson method. NewtonRaphson method gives you an iterative equation of finding . . . or iterative formula to find out the root of the equation f of x equal to 0, and it looks like this. So this is the iterative equation which you have to write, or which you have to use, to find out the root of the equation f of x equal to 0. derivative of secx tanx
So basically this recursive relationship involves that you choose some initial guess of xi and you get xsubiplus1, then you plug it back in here, you get xsubiplus2, and you continue this process until you find out that your roots are within a prespecified tolerance.
Now, you can very well see that in order to be able to use this recursive relationship, you've got to know what the derivative of the function f is.derivative of secx tanx
So that is required, as you see in the denominator there, so you have to calculate that symbolically to be able to do that.
So that's one of the drawbacks of the NewtonRaphson method, that you have to calculate the derivative of the function f symbolically to be able to use the NewtonRaphson method.
So, one of the ways to get around it is by using an approximation for the derivative of the function, and I can approximate it as the value of the function at xi minus the value of the function at xsubiminus1, divided by xi minus xsubiminus1, like this.
derivative of secx tanx
Where does this come from?
This is basically your . . . is basically your divided difference polynomial approximation of the derivative of the function, and you're using backward divided difference. So,
derivative of secx tanx
if this formula looks familiar, it is, because it the backward divided difference formula to approximate the first derivative of a function.
So what you are finding out is that, by taking this approximation, you are saying that, hey, I don't have to find the f prime, or the derivative of the function f, no longer do I have to find it exactly, I can use an approximation instead.
Now, the only thing which you are introducing which is, in addition to what we had in NewtonRaphson method, is that you're introducing another unknown here, which is xsubiminus1.derivative of secx tanx
So what that means is that you have to make two guesses now, not only will you have to guess what, or make an estimate, or have an initial guess for xi, but also an initial guess for xsubiminus1, so you will need two initial guesses to start the process, and then get a new guess.
So, if I look at . . . if I substitute this formula back into my NewtonRaphson method formula, I'll get something like this, xi minus f of xi divided by f of xi minus f of xsubiminus1, divided by xi minus xsubiminus
1. So I'm going to simplify it a little bit by taking this to the numerator, and this is what I'm going to get, I'm going to get xsubiplus1 is equal to xi minus the value of the function at xi times xi minus xsubiminus1, divided by the value of the function at xi minus the value of the function at xsubiminus1.derivative of secx tanx
And that's what the secant method is, and this is the secant method formula, so secant method formula for f of x equal to 0.
Now, what you are seeing here is that now you have to have two initial guesses, because you have xi here, xsubiminus1, so you'll have to have two initial guesses to begin with.derivative of secx tanx
So once you have chosen xsubiminus1 and xi, you'll be able to calculate this righthand side and get xsubiplus1, then you're going to use xi and xsubiplus1 to get xsubiplus2, so, again, you will continue this process until you find out that . . .derivative of secx tanx
that your root has been found within a prespecified tolerance, and that's when you will stop the algorithm.
But, again, you've got to understand that you need two initial guesses.Again, although you need two initial guesses, it does not mean that this is a . . . this is a bracketing method, this is still an open method, so secant method is still an open method, because the two initial guesses which you are going to use, they don't necessarily have to bracket the root.So, although you have two initial guesses, but they don't necessarily have to bracket the root, so it still falls under the category of the open method, just like the NewtonRaphson method.And that's the end of this segment. . . .
Post a Comment
Post a Comment