Understanding the Derivative of y = 2x sqrt(x^2 + 1)

[lotus_daemon]

y = 2x square root of (x^2+1)

I'm not exactly sure how to start off this problem. Am I supposed to use the chain rule or some other rule? I really need help on understanding this, so if anyone is kind enough to provide detailed steps, I thank you so much.

 

[VeeEight]

What other rules do you know for differentiating a function (power rule, product rule, quotient rule, etc) that would apply here?

 

[lotus_daemon]

What other rules do you know for differentiating a function (power rule, product rule, quotient rule, etc) that would apply here?

It looks like I would use the chain rule for this, but I don't know "how" to start it out. =(

 

[VeeEight]

There are powers and products of terms in the function, so why wouldn't you use the power rule and product rule? Do you know what they are and how to use them? How do they apply to this function?

 

[lotus_daemon]

There are powers and products of terms in the function, so why wouldn't you use the power rule and product rule? Do you know what they are and how to use them? How do they apply to this function?

I know what they are but I didn't think that I could apply them in this particular problem. If it's possible, can you please show me how? =( I'm used to the chain rule...

 

[VeeEight]

Why did you think that they don't apply for this problem?

You have the function written as y= 2x sqrt(x^2 + 1)
which is the same as y=2x (x^2 + 1)^(1/2)
So let 2x be some function, call it g(x) and (x^2 + 1)^(1/2) be another function, call it h(x).

You can write y as the product of two functions, y=h(x)g(x). So you can apply the product rule here. Remember to apply the power rule and chain rule when you differentiate h(x)

 

[lotus_daemon]

Why did you think that they don't apply for this problem?

You have the function written as y= 2x sqrt(x^2 + 1)
which is the same as y=2x (x^2 + 1)^(1/2)
So let 2x be some function, call it g(x) and (x^2 + 1)^(1/2) be another function, call it h(x).

You can write y as the product of two functions, y=h(x)g(x). So you can apply the product rule here. Remember to apply the power rule and chain rule when you differentiate h(x)

Hmmm...okay. I get it now. Thank you for your help. =)