Find the derivative of the function using the chain rule

In summary, the derivative of the function is x*cosx + sinx. The chain rule can be used to simplify the derivative, but it should be taken after the product rule has been used.
  • #1
frosty8688
126
0
1. Find the derivative of the function



2. [itex]\left(y= x sin\sqrt{x}\right)[/itex]



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 [itex]\sqrt{x}[/itex] is the inner function.
 
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  • #2
frosty8688 said:
1. Find the derivative of the function
2. [itex]\left(y= x sin\sqrt{x}\right)[/itex]
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 [itex]\sqrt{x}[/itex] 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: [itex]f(x) = \left( \sin \sqrt{x} \right)^2[/itex].)

So take the product rule, and then the chain rule, as you said you originally done.
 
  • #3
Ok, thanks for the advice.
 
  • #4
The answer would be [itex]\frac{x}{2sinx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]
 
  • #5
frosty8688 said:
The answer would be [itex]\frac{x}{2sinx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]
Try again! :smile:

It looks like you probably implemented the product rule correctly, but your chain rule took a wrong turn I think...
 
  • #6
It would be [itex]\frac{x}{2cosx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]
 
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  • #7
frosty8688 said:
It would be [itex]\frac{x}{2cosx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]
What is the definition of the chain rule?

Eg, if y=f(g(x)), then y'=? in terms of f, f', g, g' ?
 
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  • #8
frosty8688 said:
It would be [itex]\left.[/itex][itex]\frac{x}{2cosx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex][itex]\right.[/itex]
You're messing up the itex tags. Just use one pair, like this:
[itex]\frac{x}{2\cos x^{1/2}} \cdot \cos x + \sin x^{1/2}[/itex]
Anyway, it's still wrong. You're not taking the derivative of [itex]\sin \sqrt{x}[/itex] correctly.
 
  • #9
It would be [itex]\frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}[/itex]
 
  • #10
frosty8688 said:
It would be [itex]\frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}[/itex]
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 [itex]\frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}[/itex]
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.
[itex]\frac{x \cdot \cos \sqrt{x}}{2\sqrt{x}} + \sin \sqrt{x}[/itex]
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:
 

1. What is the chain rule?

The chain rule is a rule in calculus that allows us to find the derivative of a composite function, which is a function that is made up of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

2. When do I use the chain rule?

The chain rule is used when taking the derivative of a composite function, where the input of one function is the output of another function. It is also used when the function has nested functions, meaning one function is inside another function.

3. How do I apply the chain rule?

To apply the chain rule, first identify the outer function and the inner function. Then, take the derivative of the outer function and multiply it by the derivative of the inner function. Make sure to also multiply by the derivative of the inner function with respect to its variable, if necessary.

4. What is an example of using the chain rule?

An example of using the chain rule would be finding the derivative of the function f(x) = sin(2x). In this case, the outer function is sin(x) and the inner function is 2x. Using the chain rule, we would first take the derivative of the outer function (cos(x)) and then multiply it by the derivative of the inner function (2). This would give us the final answer of f'(x) = 2cos(2x).

5. What are common mistakes when using the chain rule?

One common mistake when using the chain rule is forgetting to multiply by the derivative of the inner function. Another mistake is not correctly identifying the inner and outer functions, which can lead to incorrect derivatives. It is important to pay attention to the order of operations and properly apply the chain rule to avoid these mistakes.

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