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AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > 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 f, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 2, x, squared. Its first derivative is f, prime, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 4, x. To find its second derivative, f, start superscript, prime, prime, end superscript, we need to differentiate f, prime. When we do this, we find that f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 6, x, plus, 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, start superscript, prime, prime, end superscript.
Leibniz's notation for second derivative is start fraction, d, squared, y, divided by, d, x, squared, end fraction. For example, the Leibniz notation for the second derivative of x, cubed, plus, 2, x, squared is start fraction, d, squared, divided by, d, x, squared, end fraction, left parenthesis, x, cubed, plus, 2, x, squared, right parenthesis.
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.(7 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)
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)
