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Second derivatives review

Review your knowledge of second derivatives.

What are second derivatives?

The second derivative of a function is simply the derivative of the function's derivative.
Let's consider, for example, the function f(x)=x3+2x2. Its first derivative is f(x)=3x2+4x. To find its second derivative, f, we need to differentiate f. When we do this, we find that f(x)=6x+4.
Want to learn more about second derivatives? Check out this video.

Notation for second derivatives

We already saw Lagrange's notation for second derivative, f.
Leibniz's notation for second derivative is d2ydx2. For example, the Leibniz notation for the second derivative of x3+2x2 is d2dx2(x3+2x2).

Check your understanding

Problem 1
f(x)=2cos(x2)
f(x)=?
Choose 1 answer:

Problem 2
d2dx2[103x3]=?
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • male robot hal style avatar for user zubair
    what does the second derivative tell us about?
    (10 votes)
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  • blobby green style avatar for user 欧子
    why the second derivative operator is not d^2y/((d^2)(x^2)), i think this way is because the product of d/dx and dy/dx is d^2y/((d^2)(x^2))
    (4 votes)
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    • male robot donald style avatar for user Venkata
      It's just a (poor and confusing) convention, but when Leibnitz first invented this notation, he thought of units of physical quantities. For example, the second derivative ($\frac{d^2y}{dx^2}$) of position is acceleration. Acceleration has the units of $\frac{m}{s^2}$. And hence, the derivative (excluding the "d" part) is also $\frac{y}{x^2}$.

      There's another thing to consider that dx isn't d times x. It isn't a product and hence, dx $\cdot$ dx can't be $d^2x^2$ (d is an operator). But, we write $d^2$ in the numerator anyway, so this kinda invalidates it.

      Honestly speaking, this is the best explanation I could find. There's no reason why it couldn't have been $\frac{d^2y^2}{dx^2}$. People used $\frac{d^2y}{dx^2}$ and we got used to it.
      (10 votes)
  • leaf green style avatar for user Yu Aoi
    is there such thing as third derivative?
    (1 vote)
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  • leafers seed style avatar for user devore.wes
    In problem #1. Why doesn't the 2 get multiplied against the entire derivative of cos(x/2)? It is only being multiplied by the first part of the chain: -sin(x/2). That doesn't seem correct.
    (1 vote)
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  • blobby green style avatar for user ZH
    I have that \dfrac{dy}{dx} = e^{5y} and I want to find \dfrac{d^2y}{dx^2}.
    So, I solve to here:
    \dfrac{d^2y}{dx^2}=5e^{5y}\cdot\dfrac{dy}{dx}
    I can substitute in e^5y for \dfrac{dy}{dx}.
    What happens if instead I divide both sides by \dfrac{dy}{dx}?
    (2 votes)
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  • duskpin ultimate style avatar for user Vanessa Slea
    Given: dy/dx = x/y. Find the 2nd derivative d2y/dx2 in terms of x and y. I found (y^2 - x^2) / y^3 and it was marked as correct. Why then do certain calculators show the answer as
    y/x ? Obviously, the calculator knows something I don't. What is it?
    (2 votes)
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  • blobby green style avatar for user tonymaione30
    hey im struggling with understanding something fundamental it seems. Im finding the derivative no problem but when i plug derivative in the numerator as a fraction using quotient rule i don't know how to evaluate it correctly what am i doing wrong?
    eg
    num- 6x^2*y-2x^3(2x^3/y) thanks for taking the time!
    den- (y)^2
    (1 vote)
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  • leaf blue style avatar for user Emily King
    If I am asked to find f'(x) of f(x)=x^4, does that mean that I am to find the second derivative (or f"(x))?
    (0 votes)
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