Find the derivative of the function using the chain rule

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Homework Help Overview

The discussion revolves around finding the derivative of the function \( y = x \sin(\sqrt{x}) \) using the chain rule and product rule in calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of the product rule combined with the chain rule, with some questioning whether the chain rule should be applied multiple times. There is also discussion about correctly identifying the outer, middle, and inner functions involved.

Discussion Status

Some participants provide feedback on the application of the product and chain rules, suggesting that initial approaches were correct while others express uncertainty about specific derivative calculations. There is a mix of attempts to clarify the correct application of these rules.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the derivative of a composition of functions and the proper use of LaTeX for mathematical expressions.

frosty8688
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1. Find the derivative of the function



2. \left(y= x sin\sqrt{x}\right)



3. I started using the product rule and then proceeded to use the chain rule, but I am wondering if I should have used the chain rule twice rather than starting with the product rule. Since I know that x is the outer function, sinx is the middle function, and \sqrt{x} is the inner function.
 
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frosty8688 said:
1. Find the derivative of the function
2. \left(y= x sin\sqrt{x}\right)
3. I started using the product rule and then proceeded to use the chain rule, but I am wondering if I should have used the chain rule twice rather than starting with the product rule. Since I know that x is the outer function, sinx is the middle function, and \sqrt{x} is the inner function.

No, you had it right the first time! The function you wrote is a product of x and a function composition (sin √x). The latter function is not inside the function x! (An example of a function composition with an outer, middle, and inner function would be something like this: f(x) = \left( \sin \sqrt{x} \right)^2.)

So take the product rule, and then the chain rule, as you said you originally done.
 
Ok, thanks for the advice.
 
The answer would be \frac{x}{2sinx^{1/2}} * cosx + sinx^{1/2}
 
frosty8688 said:
The answer would be \frac{x}{2sinx^{1/2}} * cosx + sinx^{1/2}
Try again! :smile:

It looks like you probably implemented the product rule correctly, but your chain rule took a wrong turn I think...
 
It would be \frac{x}{2cosx^{1/2}} * cosx + sinx^{1/2}
 
Last edited by a moderator:
frosty8688 said:
It would be \frac{x}{2cosx^{1/2}} * cosx + sinx^{1/2}
What is the definition of the chain rule?

Eg, if y=f(g(x)), then y'=? in terms of f, f', g, g' ?
 
Last edited by a moderator:
frosty8688 said:
It would be \left.\frac{x}{2cosx^{1/2}} * cosx + sinx^{1/2}\right.
You're messing up the itex tags. Just use one pair, like this:
\frac{x}{2\cos x^{1/2}} \cdot \cos x + \sin x^{1/2}
Anyway, it's still wrong. You're not taking the derivative of \sin \sqrt{x} correctly.
 
It would be \frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}
 
  • #10
frosty8688 said:
It would be \frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}
That's the ticket! :smile:

But to simplify further, you could cancel a sqrt(x) top and bottom from the first term.
 
  • #11
frosty8688 said:
It would be \frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}
Let's clean up the LaTeX a bit more. Use the \cdot tag instead of a "*" for multiplication. Also, when writing trig functions, put a "\" before them and a space before the variable, like \sin x.
\frac{x \cdot \cos \sqrt{x}}{2\sqrt{x}} + \sin \sqrt{x}
Anyway, you're almost done. You can simplify a little more. Notice the "x" in the numerator and the "√x" in the denomiator?EDIT: this was posted before oay's previous post was edited. :wink:
 
  • #12
eumyang said:
EDIT: this was posted before oay's previous post was edited. :wink:
Haha! :smile:

I honestly edited mine before I saw your post though... So it's a win-win! :smile:
 

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