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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 3
Lesson 8: Calculating higher-order derivativesSecond 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 . Its first derivative is . To find its second derivative, , we need to differentiate . When we do this, we find that .
Want to learn more about second derivatives? Check out this video.
Notation for second derivatives
We already saw Lagrange's notation for second derivative, .
Leibniz's notation for second derivative is . For example, the Leibniz notation for the second derivative of is .
Check your understanding
Want to try more problems like this? Check out this exercise.
Want to join the conversation?
- what does the second derivative tell us about?(6 votes)
- It tells us the rate of change of the rate of change. For example, acceleration is the second derivative of a position function, like velocity is the first derivative.(12 votes)
- 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))(3 votes)
- 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.(6 votes)
- is there such thing as third derivative?(1 vote)
- Yes, we can find any number of derivatives as long as each derivative is also differentiable.(8 votes)
- 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)
- It was multiplied by the derivative, that is why it went away. The derivative of 2cos(x/2) is 2 d/dx cos(x/2). You can use the chain rule to differentiate it, so you get 2*(1/2*-sin(x/2)). This simplifies to -sin(x/2) because 2 * 1/2 = 1. Does this help?(5 votes)
- 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) - 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)- So you have not mention what you are trying to differentiate.(1 vote)
- 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)
- If you are asked to find the first derivative (f'(x)), you are to find the first derivative (f'(x)).(0 votes)