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First derivatives

First derivatives

As you work with functions more, you’ll learn that there’s so much you can do with them! One of the first things you’ll learn is the first derivatives (yes, there are more!), so let’s just start there and make sure we have our bearings.


What does it mean to find the first derivative?

A derivative of a function is a rate of change of a function with respect to a change of a variable.

Did you know: finding the first derivative of the function $$f(x)$$ at $$x=x_0$$ is actually determining the slope of the tangent line to the graph of the function at $$x_0$$? Yep. We knew. And we love it.

To simplify the process of differentiation, we’ll use differentiation rules rather than the definition of the derivative. Here’s a table you’ll definitely want to keep handy:

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 first derivative so useful?

Functions and first derivatives are actually used all the time, from velocity to acceleration and lots of other applications. Whether it’s the rate of change or the slope of the tangent line, people are using first derivatives constantly to make our world smoother, simpler, and more predictable. Pretty cool, right?

How to find the first derivative

Now that we know why we need them, let’s learn how to find first derivatives! We’ll walk through some detailed examples together.

Example 1

Find the first-order derivative with respect to $$x$$:


Take the derivative of both sides:


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


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


Next, use $$\frac{d}{dx}(x^{n})={n}\times x^{{n}-1}$$:


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


Calculate the product:


Nice! The first order derivative with respect to $$x$$ of $$y=5x^2+3x$$ is:


Example 2

Take the first-order derivative with respect to $$x$$:

$$y^{2} = \frac{x^{2} - 49}{x^{2} + 49}$$

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

$${\frac{d}{dx}}(y^{2}) = {\frac{d}{dx}}\left(\frac{x^{2} – 49}{x^{2} + 49}\right)$$

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

$${{\frac{d}{dx}}(y^{2})\times \frac{dy}{dx}} = {\frac{d}{dx}}\left(\frac{x^{2} – 49}{x^{2} + 49}\right)$$

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

$${{\frac{d}{dx}(y^{2})}\times \frac{dy}{dx}} = {\frac{\frac{d}{dx}({x^2-49})\times {(x^2+49)}-{(x^2-49)}\times \frac{d}{dx}({x^2+49})}{{(x^2+49)}^2}}$$

Next, use $$\frac{d}{dx}(x^{n})={n}\times x^{{n}-1}$$:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{\frac{d}{dx}({x^2-49})\times {(x^2+49)}-{(x^2-49)}\times \frac{d}{dx}({x^2+49})}{{(x^2+49)}^2}}$$

Since our expression on the right side is long, let’s determine the derivative of $$x^2+49$$ and $$x^2-49$$:

$$\frac{d}{dx}({{x^2}+{49}}),~ \frac{d}{dx}({{x^2}-{49}})$$

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

$$\frac{d}{dx}{{(x^2)}+\frac{d}{dx}{(49)}},~ \frac{d}{dx}({{x^2})-\frac{d}{dx}({49}})$$

Next, use $$\frac{d}{dx}(x^{n})={n}\times x^{{n}-1}$$:

$${2x}+{\frac{d}{dx}{(49)}},~ {2x}-\frac{d}{dx}({49})$$

The derivative of any constant is equal to zero, since the derivative represents the rate of change and the constant function does not change. In other words, the derivative of $$49$$ is equal to $$0$$:

$${2x}+{0},~ {2x}-{0}$$

Remove the zero, since adding/subtracting zero doesn’t change the value:

$${2x},~ {2x}$$

Remember our equation:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{{\frac{d}{dx}({x^2-49})}\times {(x^2+49)}-{(x^2-49)}\times {\frac{d}{dx}{(x^2+49)}}}{{(x^2+49)}^2}}$$

Substitute $$\frac{d}{dx}(x^2-49)=2x$$ and $$\frac{d}{dx}(x^2+49)=2x$$ into the equation:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{{2x}\times {(x^2+49)}-{(x^2-49)}\times {2x}}{{(x^2+49)}^2}}$$

Distribute $$2x$$ through the parentheses:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{{2x}\times {x^2+{2x}\times49}-{x^2\times{2x}+49}\times {2x}}{{(x^2+49)}^2}}$$

Simplify the expression on the right side:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{ {{2x^3}+98x}{-2x^3}+{98x}}{{(x^2+49)}^2}}$$

Since two opposites add up to $$0$$, remove them from the expression:

$${{{2}y}\times \frac{dy}{dx}} = {\frac{ {98x}+{98x}}{{(x^2+49)}^2}}$$

Add like terms:

$${{2y}\times \frac{dy}{dx}} = {\frac{ {196x}}{{(x^2+49)}^2}}$$

Divide both sides of the equation by $$2y$$:

$${{2y}\times \frac{dy}{dx}}\div {2y} = \frac{196x}{{(x^2+49)}^2} \div { 2y}$$

Notice that, on the left side of the equation, we have the expression $$2y$$ divided by itself, which cancels out:

$${\frac{dy}{dx}} = \frac{196x}{{(x^2+49)}^2} \div {2y}$$

To divide a fraction by an expression, multiply by the reciprocal of that expression:

$${\frac{dy}{dx}} = \frac{196x}{{(x^2+49)}^2} \times \frac1{2y}$$

Simplify the expression on the right side:

$${\frac{dy}{dx}} = \frac{98x}{{(x^2+49)}^2} \times \frac1{y}$$

Multiply the fractions:

$${\frac{dy}{dx}} = \frac{98x}{{(x^2+49)}^2y}$$

Phew! Nice work!

When you go to try this method on other problems, remember to use these steps:

Study summary

  1. Take the derivative of each side of the equation, with respect to a variable.
  2. Use differentiation rules.
  3. Simplify the expression, if possible.
  4. Rewrite the equation by isolating the first derivative of the dependent variable.

Do it yourself!

It might not be top of your to-do list, but practicing your math skills can really help you on that next big test. If you want some extra practice on first-order derivatives, try these problems:

Take the first-order derivative with respect to $$x$$:

  1. $$2y^{3} +5x=1$$
  2. $$\ln{x} = 4y^2+3x$$
  3. $$\sqrt{y}=\frac{1}{x^2-9}$$
  4. $$e^y=x^6+3x^3$$


  1. $$\frac{dy}{dx}=-\frac{5}{6y^2}$$
  2. $$\frac{dy}{dx}=\frac{1-3x}{8xy}$$
  3. $$\frac{dy}{dx}=-\frac{4x\sqrt{y}}{x^4-18x^2+81}$$
  4. $$\frac{dy}{dx}=\frac{6x^5+9x^2}{e^y}$$

If you need some extra help on those (or any problem!), just scan it with your Photomath app, and we’ll walk you through every single step!

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