# Double derivative of x^2/(1 + x)

#### timmehk

$$\displaystyle y=x^2/(1+x)$$ Find the derivative of F"(1), By the " it means to find the double derivative, but I have no idea what to do, could anybody help me? thanks (Talking)

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#### skeeter

MHF Helper
$$\displaystyle y=x^2/(1+x)$$ Find the derivative of F"(1), By the " it means to find the double derivative, but I have no idea what to do, could anybody help me? thanks (Talking)
start by finding the first derivative ... can you do that?

#### timmehk

Ok I get $$\displaystyle 1/(1+x)^2$$ for the first derviative, but how do I solve for the second derivative? How can I use the quotient rule if I don't have an x in the numerator? Or, do I use the chain rule and disregard the fact that its division.

#### tom@ballooncalculus

Just in case a picture helps...

... shows the first derivative, where

... is the product rule, straight lines differentiating downwards. But,

... the chain rule, is involved - it hardly leaves a trace in this instance, but we can zoom in on it...

Straight continuous lines still differentiate downwards with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Writing the bottom row with a common denominator...

Then a similar routine to differentiate again...

Then plug in 1 for x.

Alternatively, you could use the quotient rule for differentiation (twice). But the pic can be useful as an overview of what's happening, and it means you don't need u and v.
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#### skeeter

MHF Helper
Ok I get $$\displaystyle 1/(1+x)^2$$ for the first derviative, but how do I solve for the second derivative? How can I use the quotient rule if I don't have an x in the numerator? Or, do I use the chain rule and disregard the fact that its division.

$$\displaystyle y = \frac{x^2}{1+x}$$

quotient rule ...

$$\displaystyle y' = \frac{(1+x)(2x) - (x^2)(1)}{(1+x)^2}$$

$$\displaystyle y' = \frac{2x + 2x^2 - x^2}{(1+x)^2}$$

$$\displaystyle y' = \frac{2x+x^2}{(1+x)^2}$$

to find the second derivative, quotient rule again ...

$$\displaystyle y'' = \frac{(1+x)^2(2+2x) - (2x+x^2) \cdot 2(1+x)}{(1+x)^4}$$

no need for simplification from this point ... since you only need to evaluate the second derivative at x = 1, just sub in 1 for x and do the arithmetic.

#### mr fantastic

MHF Hall of Fame
$$\displaystyle y=x^2/(1+x)$$ Find the derivative of F"(1), By the " it means to find the double derivative, but I have no idea what to do, could anybody help me? thanks (Talking)
Note that $$\displaystyle \frac{x^2}{1 + x} = x - 1 + \frac{1}{x + 1}$$ using polynomial long division. Getting the double derivative should now be a simple process.

tom@ballooncalculus