Calculating derivatives (2 questions)

[curlybit89]

Hello! I took an exam awhile back and missed two questions but have no feedback describing what went wrong. Let me know what you think.

Homework Statement

Derivative of x / 1 - x

Second derivative of -1/3x

2. The attempt at a solution

My answer for 1 was the number -

(1)(1-x) - (x)(-1) / (1-x)^2

1-x / (1-x)^2 - Cancel leaving x/1-x

x / -x = -1

-1 / 1 = -1

---

Second problem:

f''(x) -1/3x
-1/3x = -3x^-1
f'(x) = (-1)(-3x)^-2 = 1/3x^2
f''(x) = (-2)(3x)^-3 = 1/-6x^3

These two problems are incorrect, I'm not sure why.

 

[CaptainEvil]

Hi There

You have a sign error I noticed for #1 - the derivative is indeed (1)(1-x) - x(-1) but that is not 1-x it is 2-x

So the derivate would be (2-x)/(1-x)^2


For your second problem, your first derivative is right = 1/3x^2
now for the second, use the quotient rule again.

f''x = [0 - (1)(6x)]/[(3x^2)^2)]
the denominator there would be 9x^4, so the 2nd derivative simplifies to -2x/3x^4

 

[mathie.girl]

Hello! I took an exam awhile back and missed two questions but have no feedback describing what went wrong. Let me know what you think.

Homework Statement

Derivative of x / 1 - x

Second derivative of -1/3x

2. The attempt at a solution

My answer for 1 was the number -

(1)(1-x) - (x)(-1) / (1-x)^2

1-x / (1-x)^2 - Cancel leaving x/1-x

x / -x = -1

-1 / 1 = -1

---

Second problem:

f''(x) -1/3x
-1/3x = -3x^-1
f'(x) = (-1)(-3x)^-2 = 1/3x^2
f''(x) = (-2)(3x)^-3 = 1/-6x^3

These two problems are incorrect, I'm not sure why.

If you have $$f(x) = \frac{x}{1 - x}$$ then:

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

For the second problem, which I believe is $$g(x) = -\frac{1}{3x} = -\frac{1}{3}\ast\frac{1}{x} = -\frac{1}{3}x^{-1}$$, then you just use the chain rule to get:

$$g'(x) = -\frac{1}{3}(-1)x^{-2} = \frac{1}{3}x^{-2} \Rightarrow g''(x) = \frac{1}{3}(-2)x^{-3} = -\frac{2}{3}x^{-3}$$