Question

Can someone help me how to remember derivative shortcuts? (e^x, product rule, quotient rule, chain, etc)

Answer 1

Oh come on...you didn't like figuring it out using the definition of the derivative??

\[\large \lim_{ h \rightarrow 0} \frac{f(x+h) - f(x)}{ h}\]

Answer 2

https://en.wikipedia.org/wiki/Differentiation_rules

Answer 3

http://myhandbook.info/form_diff.html

Answer 4

you are advised to remember the standard forms for most functions.

For product rule: vu' + uv'
For quotient rule have all the v's on the outside:

\[(\frac{ u }{ v })' = \frac{ vu' - uv' }{ v^2 }\]

Chain rule: two variants exist.
If y = f(g(x))
Let u = g(x)
\[\frac{ dy }{ dx } = \frac{ dy }{ du } \times \frac{ du }{ dx }\]

Alternatively y = f(g(x))
\[y' = f'(g(x)) \times g'(x)\]
This form might be more easier to remember. Derive the 'outside' function (f) as you would but have your answer in terms of the 'inside' function (g) then derive the inside function and multiply the result together.

For example
\[y = \sin^2 4x\]
The outside function is the power (Square) and the inside is the sin of 4x. (actually this is three functions altogether if you include the 4x)
Sp we derive the outside which is a power of 4. normally power rule gives us 4x^3. Replace the x with sin 4x so your get 4 sin^3 (4x) instead. Next step you now derive the sin(4x) to get 4 cos(4x) (we use the function of function rule again in deriving the 4x to get a 4 out as well)

This can be generalised to all your function rules and becomes extremely handly when you study integration by substitution. Observe:
\[\frac{ d }{ dx } x^n = n x^{n-1} \rightarrow \frac{ d }{ dx } [f(x)]^{n} = n{f(x)}^{n-1} f'(x)\]
\[\frac{ d }{ dx } e^x = e^x \rightarrow \frac{ d }{ dx }e^{f(x)} = f'(x) e^{f(x)} \]
\[\frac{ d }{ dx } \ln(x) = \frac{ 1 }{ x } \rightarrow \frac{ d }{ dx }\ln(f(x)) = f'(x) \times \frac{ 1 }{ f(x) } = \frac{ f'(x) }{ f(x) }\]
\[\frac{ d }{ dx } \sin(x) = \cos(x) \rightarrow \frac{ d }{ dx } \sin(f(x)) = f'(x) \cos(f(x))\]
\[\frac{ d }{ dx } \tan x= \sec^2x \rightarrow \frac{ d }{ dx } \tan(f(x)) = f'(x) \sec^2(f(x))\]
\[\frac{ d }{ dx } \tan^{-1} x = \frac{ 1 }{ 1+x^2} \rightarrow \frac{ d }{ dx } \tan^{-1} f(x)) = \frac{ f'(x) }{ 1+[f(x)]^2 }\]

So really you derive almost like normal but replace x with f(x) and then multiply the result by the derivative of f.

Answer 5

Unless you want to learn how to derive formulas yourself (see johnweldon's comment, above), you'll really need to learn, practice and review those derivative formulas. One of the very few "shortcuts" I can think of is that formulas for the derivatives of a sum and of a difference are the same except for sign:

\[\frac{ d }{ dx }(a+b)=\frac{ da }{ dx }+\frac{ db }{ dx}\]

Answer 6

\[\frac{ d }{ dx}(a-b)=\frac{ db }{ dx }-\frac{ db }{ dx }\]

Answer 7

Copy down the formulas you've actually been using. When in doubt about which formula to use, refer to your list. This constant use will help you to remember the formulas.
Add to your list as you progress through you calculus course.

Answer 8

@mww thank you :) that kinda helped with my question!
@johnweldon1993 the limit derivative gets really messy for me and end up missing something but thank you though lol
@mathmale true i dont think ive been writing them down to help me remember. i always just refer to the book

Answer 9

My point was that the more you see and handle these derivative formulas, the more likely you are to remember them when you need them. But I applaud your referring to your textbook.