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Second derivatives

Second derivatives

Oh boy… it’s time to start graphing more complicated functions.

Don’t sweat it! You can handle it, thanks to derivatives.

We already learned that the first derivations can tell us if the functions are increasing or decreasing at certain intervals. The second derivation can tell us a little bit more, specifically the concavity of the function at a certain interval.

Let’s dive in!

What does it mean to find the second derivative?

A derivative of a function is a rate of change of a function with respect to a change of a variable. In fact, to find the second derivative of the function f(x) at x=x0 means to determine if the slope of the tangent line is increasing or decreasing.

To simplify the process of differentiation, we use differentiation rules rather than the definition of the derivative. Here are the rules you’ll be using:

Constant multiple property of derivatives ddx(c×f(x))=c×ddx(f(x))
Sum rule for derivatives ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x))
Difference rule for derivatives ddx(f(x)g(x))=ddx(f(x))ddx(g(x))
Product rule for derivatives ddx(f(x)×g(x))=ddx(f(x))×g(x)+f(x)×ddx(g(x))
Quotient rule for derivatives ddx(f(x)g(x))=ddx(f(x))×g(x)f(x)×ddx(g(x))(g(x))2
The Chain rule (fg)(x)=f(g(x))×g(x) or dydx=dydududx when y=f(u),u=g(x)
Derivative of the inverse function (f1)(x)=1f(f1(x))

Why is the second derivative so useful?

The second derivative can be really useful if we want to graph a function.

If the second derivative of a point at a certain interval is negative, the function is concave down on that interval; if it’s negative, the function is concave up at that interval. When the second derivative of a point is equal to zero, that point can be an inflection point.

Trust us, you’ll want to know that!

How to find the second derivative

Now that we know why we use the second derivative, let’s see how we find one!

Example 1


Find the second derivative with respect to x:

y+3=2x2+1

Take the derivative of each term, with respect to x:

ddx(y+3)=ddx(2x2+1)

Use the differentiation rule ddx(f+g)=ddx(f)+ddx(g):

ddx(y)+ddx(3)=ddx(2x2)+ddx(1)

The derivative of a constant is always zero, so substitute 0 for ddx(3) and ddx(1):

ddx(y)+0=ddx(2x2)+0

Removing zero doesn’t change the value, so remove it from the expressions on the left and right side:

ddx(y)=ddx(2x2)

Use the chain rule ddx(y)=ddy(y)×dydx:

ddy(y)×dydx=ddx(2x2)

Use the differentiation rule ddx(a×f)=addx(f):

ddy(y)×dydx=2ddx(x2)

The derivative of a variable to the first power is always 1:

1×dydx=2ddx(x2)

Use the differentiation rule ddxxn=nxn1:

1×dydx=2×2x

Calculate the products:

dydx=4x

To find the second derivative, take the derivative of each term, with respect to x:

ddx(dydx)=ddx(4x)

Use the differentiation rule ddx(ddx(y))=d2ydx2:

d2ydx2=ddx(4x)

Use the differentiation rule ddx(a×x)=a:

d2ydx2=4

Nice! The second derivative of y+3=2x2+1 is:

d2ydx2=4

Example 2


Find the second derivative with respect to x:

Take the derivative of both sides:

ddx(x2y2)=ddx(36)

Use the differentiation rule ddx(f+g)=ddx(f)+ddx(g):

ddx(x2)ddx(y2)=ddx(36)

Use the differentiation rule ddxxn=nxn1:

2xddx(y2)=ddx(36)

Use the chain rule ddx(y2)=ddy(y2)×dydx:

2xddy(y2)×dydx=ddx(36)

The derivative of a constant is always 0:

2xddy(y2)×dydx=0

Use the differentiation rule ddxxn=nxn1:

2x2y×dydx=0

Move the variable 2x to the right-hand side and change its sign:

2y×dydx=2x

Divide both sides of the equation by 2y to isolate dydx on one side:

dydx=xy

To find the second derivative, take the derivative of each term, with respect to x:

ddx(dydx)=ddx(xy)

Use the differentiation rule ddx(ddx(y))=d2ydx2:

d2ydx2=ddx(xy)

Use the differentiation rule ddx(fg)=ddx(f)×gf×ddx(g)g2:

d2ydx2=ddx(x)×yx×ddx(y)y2

The derivative of a variable to the first power is always 1:

d2ydx2=1×yx×ddx(y)y2

Use the chain rule ddx(y)=ddy(y)×dydx to find the derivative:

d2ydx2=1×yx×ddy(y)×dydxy2

The derivative of a variable to the first power is always 1:

d2ydx2=1×yx×1×dydxy2

Any expression multiplied by 1 remains the same:

d2ydx2=yx×dydxy2

Use the equation dydx=xy to substitute xy for dydx:

d2ydx2=yx×xyy2

Simplify the expression on the right side:

d2ydx2=y2x2y3

We did it! The second derivative of x2y2=36 is:

d2ydx2=y2x2y3

That wasn’t so bad, right? Now that we’ve walked through some detailed examples, let’s review the overall process so you can learn how to use it with any problem:

Study summary

  1. Take the derivative of each term, with respect to a variable.
  2. Use the differentiation rules.
  3. Simplify the expression, if possible.
  4. Rewrite the equation by isolating the first derivative of the dependent variable.
  5. Again, take the derivative of each term, with respect to a variable.
  6. Use the differentiation rules.
  7. Simplify the expression, if possible.
  8. Rewrite the equation by isolating the second derivative of the dependent variable.

Do it yourself!

Feeling good? Feeling nervous? Either way, it’s a good idea to practice! Here are some problems that can help you commit that method to your skillset:

Find the second derivative with respect to x:

  1. y3=5x3
  2. x3=5x2y3
  3. y=x2
  4. ey=3x2+1

Solutions:

  1. d2ydx2=509y5
  2. d2ydx2=30y318xy3200x2+120x318x49y5
  3. d2ydx2=4y+8x2yy
  4. d2ydx2=6ey36x2e2y

If you’re struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you get stuck or lost, scan the problem using your Photomath app, and we’ll walk you through!

Here’s a sneak peek of what you’ll see: