Find the derivative of the function using the chain rule

[frosty8688]

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.

 

[eumyang]

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.

 

[frosty8688]

Ok, thanks for the advice.

 

[frosty8688]

The answer would be [itex]\frac{x}{2sinx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]

 

[skiller]

The answer would be [itex]\frac{x}{2sinx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]

Try again!

It looks like you probably implemented the product rule correctly, but your chain rule took a wrong turn I think...

 

[frosty8688]

It would be [itex]\frac{x}{2cosx^{1/2}}[/itex] * cosx + sinx[itex]^{1/2}[/itex]

 

[skiller]

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' ?

 

[eumyang]

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.

 

[frosty8688]

It would be [itex]\frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}[/itex]

 

[skiller]

It would be [itex]\frac{x*cosx^{1/2}}{2x^{1/2}} + sinx^{1/2}[/itex]

That's the ticket!

But to simplify further, you could cancel a sqrt(x) top and bottom from the first term.

 

[eumyang]

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.

 

[skiller]

EDIT: this was posted before oay's previous post was edited.

Haha!

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