Question

I need explanations on how to take the derivative of a trig functions. Could somebody explain how to do so and then help me work out a few examples?

Answer 1

Sure, with basic trig functions. Learning the derivatives is simple memorization or using a cheat sheet.

\[\frac{d}{dx} \sin x = \cos x\]
\[\frac{d}{dx} \cos x = -\sin x\]
\[\frac{d}{dx} \tan x = \sec^2 x\]

Answer 2

And for csc, sec, cot?

Answer 3

they can be found by writing in terms of sin,cos,or tan , Then use the chain rule.

Anyway here are the derivatives....you can work them out on your ow.
\[\frac{d}{dx} \csc x = -\csc x \cot x\]
\[\frac{d}{dx} \sec x = \sec x \tan x\]
\[\frac{d}{dx} \cot x = -\csc^2 x\]

Answer 4

So, if I'm taking the derivative of a trig function, do I basically just plug that in in place of it? So if I need to take the derivative of lets say, 3cosx, it would just be -3sinx?

Answer 5

correct, thats it..
However if its cos(3x), you must use chain rule ----> -3sin(3x)

Answer 6

Can you explain the chain rule to me a bit more in detail? Someone helped me by using it in one of my last questions, but I don't *fully* understand it. I know you multiply whatever is inside of the parenthesis at the end (i don't know how to word it). So in any situation, if something is being multiplied by something in parenthesis, do I just multiply it at the end in the derivative?

Answer 7

yes thats pretty much the right idea.

lets do an example
\[\sin^2 x\]
So you want to think of 2 functions, theres the sin function and the squared function.
\[ u = \sin x \rightarrow f(u) = u^2\]
chain rule says:
\[f'(u) = 2u * du\]
where du is derivative of "u" function, in this case sin x
du = cos x

\[f'(u) = 2u*du = 2\sin x \cos x\]

Answer 8

Okay, and will the chain rule ALWAYS say 2u*du? Or is that specific for this problem? Would it be 3u*du if the example was sin^3(x)?

Answer 9

Another one:
\[\sin(3x^2 -1)\]
In this case, let
\[ u = 3x^2 - 1 \rightarrow f(u) = \sin u\]
\[f'(u) = \cos u * du\]
\[du = 6x\]
\[f'(u) = \cos(3x^2 -1) (6x)\]

Answer 10

It depends on the function. you have the right idea for sin^3, however it would be 3u^2 du

Answer 11

It's making a lot more sense. Now, how do I determine which part of the function is u and which part is f(u)? Is u always the part of the function that is being altered? Like, in the first example, u was the term that was being squared.

Answer 12

right thats the trick . I like to think of the "u" function as being the inner function
While f(u) is the outer base function

Answer 13

\[\frac{1}{x^3 -1}\]
what is u?

Answer 14

u= x^3-1 and f(u) = 1/u

Is that right?

Answer 15

yes, good job

Answer 16

sorry did you have your own questions?

Answer 17

Yes, I have a couple if you have the time to help me through them.

Answer 18

sure

Answer 19

Okay. So this is one of the simpler ones.

y=8xsinx

So, I know the derivative of sinx is cosx.

Answer 20

Would u be sinx and f(u) be 8x(u)?

Answer 21

note, f(u) cannot have an "x" in it

This looks more like a problem for product rule

Answer 22

the u, f(u) notation only works where chain rule applies

Answer 23

Is it just a matter of recognizing if chain rule would work for a function or is there some sort of process I can do to check if the function would require chain rule?

Answer 24

yep, first check to see if you only have base functions or if its a more complicated function.
in this case you just have 2 base functions multiplied together

Answer 25

What's the product rule? I know the answer to the equation is 16cosx-8xsinx

Answer 26

product rule
\[(fg)' = f'g + fg'\]
f = 8x, g = sin x

Answer 27

your answer is wrong, are you sure its 8x*sin(x) ?

Answer 28

OH wait! y' means the first derivative, and y'' means second derivative, correct?

Answer 29

If so, then I'm needing to find the second derivative.

Answer 30

Of y=8xsinx

Answer 31

ahh, yes y'' means 2nd derivative

Answer 32

Sorry about that! So to find second derivatives, is it easier to find the first derivative and then take the derivative of that? Or is there another process that I should be doing?

Answer 33

unfortunately there's no shortcut. You have to find y' first . Then take derivative again

Answer 34

Okay. So for the first derivative, it would be 8sinx+8x(cosx), right?

Answer 35

correct

Answer 36

So how do I go about finding the derivative of that? Do I have to simplify 8x(cosx) if possible?

Answer 37

The derivative of 8sinx would be 8cosx, right? And then the derivative of 8x(cosx)...would just be 8xsinx?

Answer 38

you take derivative of each term
You will have to use product rule again for the "8xcosx" term

Answer 39

Okay, so the derivative of 8sinx is 8cosx.

And the derivative of "8xcosx" would be 8(sinx)+8x(sinx) ?

Answer 40

8(cosx)+8x(sinx) **

Answer 41

almost, remember derivative of cos = -sin

Answer 42

ohhh, that's right. So 8(cosx)+8x(-sinx) ?

Answer 43

So in all, I would have 8cosx+8cosx+8x(-sinx) ? The two "8cosx" would add to 16cosx+8x(-sinx). And then how do I go further than that?

Answer 44

you got it. you cant simplify it any further

Answer 45

It's saying the correct answer is 16cosx-8xsinx though, would I need to factor out the - from the sinx?

Answer 46

yes

Answer 47

So it's done! Awesome! Do you want to help me out on one of the more complicated ones? (At least they seem more complicated to me)

Answer 48

sure

Answer 49

Find dy/dt.

y=cos^6(pi(t)-9)

Answer 50

\[y=\cos^6(\pi(t)-9)\])

Answer 51

First, is there a difference between finding y' and finding dy/dt?

Answer 52

no they are the same thing.
So this is a nested function. We need to use chain rule twice.

Answer 53

Okay, so I'm guessing I do the chain rule in the parenthesis first. The pi(t)-9

Answer 54

And u would be...maybe pi(t) and f(u) = u-9?

Answer 55

not quite, that makes f(u) too simple
try u = pi(t) - 9

Answer 56

So f(u) = cos^6(u)?

Answer 57

yes

Answer 58

Okay. And the derivative of pi(t)-9 would be pi and the derivative of cosu would be -sinu...so it would be -sin(pi) ?

Answer 59

wait no it would be

Answer 60

-6sin^5(pi) ?

Answer 61

-6sin^5(pi(t)-9)(pi) ?

Answer 62

getting closer :)

Important when dealing with trig functions, the derivative will ALWAYS keep the same input in parenthesis as original
In this case --->( pi(t) - 9)

The derivative du = pi is just multiplied on the outside.

Ok this is where i said we need to do chain rule twice
\[\cos^6 u\]
\[w = \cos u \rightarrow f(w) = w^6\]
\[f'(w) = 6w^5 * dw*du\]
\[= 6 \cos^5 u (-\sin u) (\pi)\]

Answer 63

Final step is plug back in (pi)t - 9 for u

Answer 64

And I can factor out that -1, so I could have -6(pi)cos^5(pi(t)-9)(sin(pi(t)-9)

Answer 65

Let me write it in the equation thing, one sec.

Answer 66

\[-6(\pi)\cos^5((\pi)(t)-9)(\sin((\pi)(t)-9)\]

Answer 67

very good
This is called a nested function where you have to use chain rule multiple times. It can take getting used too

Answer 68

And how did you determine that it was a nested function? Is it because the cos had an exponent? So if it was just cos(pi(t)-9), it wouldn't be a nested function?

Answer 69

correct

Answer 70

Jesus, this is a lot to take in!

Answer 71

yes, welcome to the world of calculus. But hopefully you have weeks to learn, is this h.s or college?

Answer 72

College.

Answer 73

Would you like to explain implicit differentiation? I have a few examples of that I need to complete.

Answer 74

sure, is this an online class? you seem to be covering multiple sections at once

Answer 75

Yes it is. I'm covering many, many sections at once.

Answer 76

thats typically not the best way to learn math , oh well

Answer 77

For the upcoming test, I have everything that you've covered with me so far. As well as Continuous/differentiable points on a graph, finding speed/acceleration/velocity of an equation, graphing derivatives, implicit differentiation, tangents to a curve, slopes of a curve, how fast a wheel is rolling...how fast water is being drained from a container..

Answer 78

My professor kind of dumps like 12-15 assignments online and then gives us a due date for them and then announces the test later along the line.

Answer 79

ok well for implicit, you need to know the product rule really well
Also just remember that only the "y" term derivative will be multiplied by dy/dx

Example:
\[xy - x^2 = 1\]
Take derivative of each term
\[(xy)' = (1)y + x(1)\frac{dy}{dx}\]
\[(-x^2)' = -2x\]
\[1' = 0\]
\[\rightarrow (y +x \frac{dy}{dx}) -2x = 0\]
Then you have to algebraically solve for "dy/dx"
\[\frac{dy}{dx} = \frac{2x-y}{x}\]

Answer 80

heres a good website with lots of notes that will help you.

http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx