Did i find the derivative correctly?

[dejan]

Hi there,
Just wondering if i found dy/dx of (x+1)^2/1+x^2 = 2(x+1)/2x and also the second derivative d^2 y/dx^2=(x+1)/2
Is that correct?? I don't think it is.

 

[d_leet]

Hi there,
Just wondering if i found dy/dx of (x+1)^2/1+x^2 = 2(x+1)/2x and also the second derivative d^2 y/dx^2=(x+1)/2
Is that correct?? I don't think it is.

No its not the derivative of a quotient is not equal to the quotient of the derivatives. Have you learned the quotient rule for differentiating these kinds of functions?

 

[dejan]

Aww i have to use the quotient rule?? *sigh*
Well we didn't go that far:( More work! Thanks anyway!

 

[Curious3141]

Aww i have to use the quotient rule?? *sigh*
Well we didn't go that far:( More work! Thanks anyway!

Well, you don't *have* to. You can use one of multiple methods,

for example, partial fraction decomposition into elementary functions with complex factors in the denominator.

Or even substitute $$x = tan \theta$$ simplify with trig identities first, then observe that $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

There are many way to skin a cat.

 

[dav2008]

Have you learned the product rule? The quotient rule is just a special case of the product rule.

Your function is a product of $$(x+1)^2$$ and $$(1+x^2)^{-1}$$

 

[dejan]

Yeah we've learned the product rule, chain and all...just that we have to graph that, but doing it without a graphics calculator.
I think i get it now.