Derivatives of expressions
Are functions your friend yet? How about derivatives?
Did you know that, sometimes, a function can be represented as only the expression? Well, they can — and guess what else? That can make it easier find the derivative.
Let’s learn how!
What does it mean to find the derivative of an expression?
To find the derivative of an expression is essentially the same as finding the derivative of a function. Sometimes we can simplify finding the derivative by removing $$f(x)$$ and just finding the derivative of the expression that’s left.
A derivative of a function is a rate of change of that function with respect to a change in variable.
The more you know: finding the first derivative of the function $$f(x)$$ at $$x_0$$ actually means finding the slope of the tangent line to the graph of the function at $$x_0$$.
To simplify the process of differentiation, we’ll use differentiation rules rather than the definition of the derivative. We’ve got the list for you right here — and trust us, you’ll need it!
| 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 derivative so useful?
Functions are endlessly important, which means, by extension, derivatives are, too! Because they tell us the rate of change, derivatives are super important for physicists, economists, and so many more people in our world (maybe even more than you realize!).
How to find the derivative of an expression
Okay, let’s talk about how to find the derivative of an expression. If you’ve already learned how to find the derivative of a function, this will probably look pretty familiar:
Example 1
Find the derivative of the expression:
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}(x^{2}) + \frac{d}{dx}(3x)$$
Next, use $$\frac{d}{dx}(x^{n})={n}\times x^{{n}-1}$$:
$${2}x^{1} + \frac{d}{dx}(3x)$$
Any expression raised to the power of $$1$$ equals itself, so:
$${2}x + \frac{d}{dx}(3x)$$
Use $$\frac{d}{dx} (a\times x)=a$$:
$${2}x + 3$$
There we go! The derivative of the expression $$x^{2} + 3x$$ is:
See? Not so bad. Let’s do one more!
Example 2
Find the derivative of the expression:
Use the differentiation rule $$\frac{d}{dx}({f}\times {g})=\frac{d}{dx}({f})\times {g}+{f}\times\frac{d}{dx}({g})$$:
$$\frac{d}{dx}{(\ln{x})}\times {x}+{\ln{x}}\times\frac{d}{dx}{(x)}$$
Use $$\frac{d}{dx} (\ln{x})=\frac1x$$:
$${\frac1x}\times {x}+{\ln{x}}\times{\frac{d}{dx}{(x)}}$$
Next, use $$\frac{d}{dx}(x)=1$$:
$${\frac1{x}}\times {x}+{\ln{x}}\times{1}$$
Simplify the expression:
$${1}+{\ln{x}}$$
We did it again! The derivative of an expression $$\ln{x}\times x$$ is:
As you can see, it’s not so different from taking the derivative of a function. To review, here’s the process you should follow for finding the derivative of an expression:
Study summary
- Use the differentiation rules.
- Find the derivative of each term.
- Simplify the expression.
Do it yourself!
Want some more practice? Try your hand at these problems and let us know if you get stuck:
Take the derivative of an expression:
- $$\frac{d}{dx}\left(e^x + 5x\right)$$
- $$\frac{d}{dx}\left(\ln{(x)}+5^x\right)$$
- $$\frac{d}{dx}\left(\sqrt{x^4-1} \right)$$
- $$\frac{d}{dx}\left(\frac{x-1}{x^2-4}\right)$$
Solutions:
- $$e^x+5$$
- $$\frac1x+\ln{5}\times5^x$$
- $$\frac{2x^3}{\sqrt{x^4-1}}$$
- $$\frac{-x^2+2x-4}{(x^2-4)^2}$$
How did you do? Do you want to check your work? Scan the problem using your Photomath app to see detailed step-by-steps of the solving method.
Here’s a sneak peek of what you’ll see:
