Finding nth Derivative of Fractional Equation x/(1+x)?

[LilTaru]

Homework Statement

Find a formula for the nth derivative of y = x/(1 + x).

Homework Equations

I know the general formula for finding the nth derivative of xⁿ.

The Attempt at a Solution

I don't know how to apply the nth derivative to a fractional equation or even if I were to rewrite y as y = x * (1 + x)^-1, I still don't know how to find the nth derivative?

 

[stevenb]

Homework Statement

Find a formula for the nth derivative of y = x/(1 + x).

Homework Equations

I know the general formula for finding the nth derivative of xⁿ.

The Attempt at a Solution

I don't know how to apply the nth derivative to a fractional equation or even if I were to rewrite y as y = x * (1 + x)^-1, I still don't know how to find the nth derivative?

Try taking the first derivative, then take the second derivative, then take the third. Do you see a pattern emerging? Can you write a general equation that specifies the nth derivative directly?

 

[LilTaru]

I am starting to sort of see a pattern emerging, but I don't know how to prove it in a general equation... I am starting to see that fⁿ(x) = something/(g(x))^n*2, but I don't see a pattern in the numerator...

I got y¹ = 2x - 1/(1 + x)²
and y² = -2x - 4/(1 + x)⁴

 

[stevenb]

I am starting to sort of see a pattern emerging, but I don't know how to prove it in a general equation... I am starting to see that fⁿ(x) = something/(g(x))^n*2, but I don't see a pattern in the numerator...

I got y¹ = 2x - 1/(1 + x)²
and y² = -2x - 4/(1 + x)⁴

Hmmm, that doesn't look right. Take the first derivative again using the product rule or the quotient rule.

http://tutorial.math.lamar.edu/Classes/CalcI/ProductQuotientRule.aspx

 

[HallsofIvy]

Well, there is your first problem- you have even the first derivative wrong. Using the quotient rule, the derivative of
y= \frac{x}{1+ x}
is
y'= \frac{(x)'(1+ x)- (x)(1+x)'}{(1+ x)^2}= \frac{1+ x- x}{(1+ x)^2}= \frac{1}{(1+ x)^2}
Writing that as (1+ x)^{-2}, y''= -2(1+ x)^{-3}, etc.

 

[LilTaru]

Okay I tried again with the product rule and got:

y¹ = -x + 1/(1 + x)²

 

[LilTaru]

Wow... apparently I have issues getting the derivative...

 

[HallsofIvy]

Again, no. If you showed each step, we might be able to tell you exactly what you are doing wrong.

 

[stevenb]

Okay I tried again with the product rule and got:

y¹ = -x + 1/(1 + x)²

No, that's not it either.

OK, let's get the process down first.

The product rule says take the first times the derivative of the second plus the second times the derivative of the first.

The two functions are

x and 1/(1+x)

The derivatives of those functions are

1 and -1/(1+x)²

Make sure you understand up to this point and then try to work out the answer for the first derivative.

 

[LilTaru]

Okay! Tried it again after refreshing on how to get a derivative... and I got:

yⁿ = -2(n - 1)(1 + x)^{-2 - (n - 1)}

 

[LilTaru]

I figured out what I did wrong... I was flipping the quotient rule... meaning I took:
f(x)g¹(x) - f¹(x)g(x) instead of the other way around

 

[stevenb]

Okay! Tried it again after refreshing on how to get a derivative... and I got:

yⁿ = -2n(1 + x)^{-2 - n}

Oh boy. You're making progress but you jumped too fast. Go slowly. Since you are having some issue with basic derivatives. Carefully write out the first through sixth derivative and you'll see that your formula doesn't have the right form.

 

[Mentallic]

Well testing your formula for n=1 gives \frac{dy}{dx}=\frac{-2}{(1+x)^3}

Which isn't right. Show us what you did, step by step, to find the first derivative.

 

[LilTaru]

Okay, just realized how wrong I had it and here is my step by step now:

y¹ = (1 + x)(1) - x(1)/(1 + x)²
= 1/(1 + x)²
= (1 + x)^-2

y² = -2(1 + x)^-3
y³ = 6(1 + x)^-4
y⁴ = -24(1 + x)^-5 and so on...

Hopefully I finally got it and now can determine the pattern?

 

[stevenb]

Okay, just realized how wrong I had it and here is my step by step now:

y¹ = (1 + x)(1) - x(1)/(1 + x)²
= 1/(1 + x)²
= (1 + x)^-2

y² = -2(1 + x)^-3
y³ = 6(1 + x)^-4
y⁴ = -24(1 + x)^-5 and so on...

Hopefully I finally got it and now can determine the pattern?

That looks good. As a reward for your diligence, I'll remind you of the factorial function

n!=n(n-1)*(n-2)*(n-3) ... 3*2*1

It should help in your final formula.

 

[LilTaru]

Oh! Thank you! I was just looking at the derivatives going where have I seen this before! Thank you to everyone for their help! It is really appreciated!

 

[LilTaru]

Alright! Fingers crossed! I got a final answer of:

yⁿ = [-(-n!)] * (1 + x)^{-2 - (n -1)}

 

[stevenb]

Alright! Fingers crossed! I got a final answer of:

yⁿ = [-(-n!)] * (1 + x)^{-2 - (n -1)}

OK, if you simplify this you would get the following.

yⁿ = n! * (1 + x)^{-(1 + n)}

This is close, but you are not getting the alternating signs from this formula.

How can you make a formula which depends on n and gives alternating signs?

 

[LilTaru]

I thought the -(-n!) would do this since:
-3! would be -3 * -2 * -1 which equals -6 and thus
-(-6) equals six... this would alternate between the signs... I can't think of another way...

 

[stevenb]

I thought the -(-n!) would do this since:
-3! would be -3 * -2 * -1 which equals -6 and thus
-(-6) equals six... this would alternate between the signs... I can't think of another way...

OK, the usual way is to write (-1)ⁿ. If this gives you the wrong sign, you can instead write -(-1)ⁿ or (-1)ⁿ⁺¹.

The use of (-n!) is not an accepted notation, as far as I know. It's better to use established conventions, but even though it's often frowned upon, one is always free to invent notation as long as you clearly define it.

 

[Mentallic]

And not just that it's not accepted notation, but what you mean would rather be (-n)! instead of -n! since the first would most likely mean (-n)(-(n-1))(-(n-2))...(-3)(-2)(-1) but the second means -(n(n-1)(n-2)...(3)(2)(1))=-(n!)