# 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 $$x=x_0$$ 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 | $$\frac{d}{dx}\left(c\times f(x)\right)=c\times\frac{d}{dx}\left(f(x) \right)$$ |

Sum rule for derivatives | $$\frac{d}{dx}\left(f(x) + g(x)\right)=\frac{d}{dx}\left( f(x) \right)+\frac{d}{dx}\left( g(x) \right)$$ |

Difference rule for derivatives | $$\frac{d}{dx}\left(f(x) - g(x)\right)=\frac{d}{dx}\left( f(x) \right)-\frac{d}{dx}\left( g(x) \right)$$ |

Product rule for derivatives | $$\frac{d}{dx}\left(f(x)\times g(x)\right)=\frac{d}{dx}\left( f(x) \right)\times g(x)+f(x)\times\frac{d}{dx}\left( g(x) \right) $$ |

Quotient rule for derivatives | $$\frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{\frac{d}{dx}\left(f(x) \right)\times g(x)-f(x)\times\frac{d}{dx}\left( g(x) \right)}{(g(x))^{2}}$$ |

The Chain rule | $$(f\circ g)^{\prime}(x)=f^{\prime}(g(x))\times g^{\prime}(x) \text{ or }\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} \text{ when }y=f(u), u=g(x)$$ |

Derivative of the inverse function | $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)}$$ |

### 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** $$x$$**:**

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

$$\frac{d}{dx}(y+3)=\frac{d}{dx}(2x^2+1)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(y)+\frac{d}{dx}(3)=\frac{d}{dx}(2x^2)+\frac{d}{dx}(1)$$

The derivative of a constant is always zero, so substitute $$0$$ for $$\frac{d}{dx}(3)$$ and $$\frac{d}{dx}(1)$$:

$$\frac{d}{dx}(y)+0=\frac{d}{dx}(2x^2)+0$$

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

$$\frac{d}{dx}(y)=\frac{d}{dx}(2x^2)$$

Use the chain rule $$\frac{d}{dx} (y)=\frac{d}{dy}(y)\times\frac{dy}{dx}$$:

$$\frac{d}{dy}(y)\times\frac{dy}{dx}=\frac{d}{dx}(2x^2)$$

Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:

$$\frac{d}{dy}(y)\times\frac{dy}{dx}=2\frac{d}{dx}(x^2)$$

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

$$1\times\frac{dy}{dx}=2\frac{d}{dx}(x^2)$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$1\times\frac{dy}{dx}=2\times2x$$

Calculate the products:

$$\frac{dy}{dx}=4x$$

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

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(4x)$$

Use the differentiation rule $$\frac{d}{dx}\left(\frac{d}{dx}(y)\right)=\frac{d^2y}{dx^2}$$:

$$\frac{d^2y}{dx^2}=\frac{d}{dx}(4x)$$

Use the differentiation rule $$\frac{d}{dx}(a\times x)=a$$:

$$\frac{d^2y}{dx^2}=4$$

Nice! The second derivative of $$y+3=2x^2+1$$ is:

### Example 2

**Find the second derivative with respect** **to **$$x$$**:**

Take the derivative of both sides:

$$\frac{d}{dx}(x^{2} – y^{2}) = \frac{d}{dx}(36)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(x^2)-\frac{d}{dx}(y^2)=\frac{d}{dx}(36)$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$2x-\frac{d}{dx}(y^2)=\frac{d}{dx}(36)$$

Use the chain rule $$\frac{d}{dx} (y^2)=\frac{d}{dy}(y^2)\times\frac{dy}{dx}$$:

$$2x-\frac{d}{dy}(y^2)\times\frac{dy}{dx}=\frac{d}{dx}(36)$$

The derivative of a constant is always $$0$$:

$$2x-\frac{d}{dy}(y^2)\times\frac{dy}{dx}=0$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$2x-2y\times\frac{dy}{dx}=0$$

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

$$-2y\times\frac{dy}{dx}=-2x$$

Divide both sides of the equation by $$-2y$$ to isolate $$\frac{dy}{dx}$$ on one side:

$$\frac{dy}{dx}=\frac{x}{y}$$

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

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{x}{y}\right)$$

Use the differentiation rule $$\frac{d}{dx}\left(\frac{d}{dx}(y)\right)=\frac{d^2y}{dx^2}$$:

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{x}{y}\right)$$

Use the differentiation rule $$\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{\frac{d}{dx}({f})\times {g}-{f}\times \frac{d}{dx}({g})}{{g}^2}$$:

$$\frac{d^2y}{dx^2}=\frac{\frac{d}{dx}(x)\times y-x\times\frac{d}{dx}(y)}{y^2}$$

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

$$\frac{d^2y}{dx^2}=\frac{1\times y-x\times\frac{d}{dx}(y)}{y^2}$$

Use the chain rule $$\frac{d}{dx}(y)=\frac{d}{dy}(y)\times \frac{dy}{dx}$$ to find the derivative:

$$\frac{d^2y}{dx^2}=\frac{1\times y-x\times\frac{d}{dy}(y)\times \frac{dy}{dx}}{y^2}$$

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

$$\frac{d^2y}{dx^2}=\frac{1\times y-x\times1\times \frac{dy}{dx}}{y^2}$$

Any expression multiplied by $$1$$ remains the same:

$$\frac{d^2y}{dx^2}=\frac{y-x\times \frac{dy}{dx}}{y^2}$$

Use the equation $$\frac{dy}{dx}=\frac{x}{y}$$ to substitute $$\frac{x}{y}$$ for $$\frac{dy}{dx}$$:

$$\frac{d^2y}{dx^2}=\frac{y-x\times \frac{x}{y}}{y^2}$$

Simplify the expression on the right side:

$$\frac{d^2y}{dx^2}=\frac{y^2-x^2}{y^3}$$

We did it! The second derivative of $$x^{2} – y^{2} = 36$$ is:

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

- Take the derivative of each term, with respect to a variable.
- Use the differentiation rules.
- Simplify the expression, if possible.
- Rewrite the equation by isolating the first derivative of the dependent variable.
- Again, take the derivative of each term, with respect to a variable.
- Use the differentiation rules.
- Simplify the expression, if possible.
- 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** $$x$$**:**

- $$y^{3}=5x-3$$
- $$x^3= 5x^2-y^3$$
- $$\sqrt{y}=x^2$$
- $$e^y=3x^2+1$$

*Solutions:*

- $$\frac{d^2y}{dx^2}=-\frac{50}{9y^5}$$
- $$\frac{d^2y}{dx^2}=\frac{30y^3-18xy^3-200x^2+120x^3-18x^4}{9y^5}$$
- $$\frac{d^2y}{dx^2}=\frac{4y+8x^2\sqrt{y}}{\sqrt{y}}$$
- $$\frac{d^2y}{dx^2}=\frac{6e^y-36x^2}{e^{2y}}$$

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:**