Question

The first derivative of sin x = -cos x. Right?

Answer 1

(sinx)'=cosx

Answer 2

yelp

Answer 3

(cosx)'= - sinx

Answer 4

so they are just their inverses?

Answer 5

what do you mean by inverse?

Answer 6

just what is the inverse of the first derivative of sin x? ....my apologies, im really tired, forgive me. a lot of calculus came togehter in my head in the last 15 hours (exam in 5 more) and I am trying my best to not get frustrated and stay straight on them...so im asking silly things here I should know. : ) lol

Answer 7

so what is the first derivative of the inverse of sin(x)? I guess is what I am asking lol

Answer 8

\[\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}\]

Answer 9

lol silly doubts are good....

Answer 10

okay, thats not what I was thinking it was going to be...haha...I heard, and I could have misheard it in class, or somewhere, but that f ' sin = cos did I hear that right and f ' cos = sin.

Answer 11

wait, sin x = -cos x. that would be its inverse, right?

Answer 12

(i am so sorry lol)

Answer 13

\[\frac{d}{dx}\sin x = \cos x \]

Answer 14

yeah, that! thats it! awesome! thats beautiful right there Ishann94 thank you!!!!

Answer 15

the (inverse of sine) x = y
arcsin(x) = y
x =sin(y)

Answer 16

:)

Answer 17

so, what does that mean exactly, since I was confusing it with inverses?

Answer 18

So: f "(x) where f(x)= x cos x is...well a mess lol

Answer 19

guess i could throw the product rule at it and run away

Answer 20

Draw a graph of both sin x and cos x of these and you'll see very clearly what everyone is saying: the the slope of the tangents to sin x is given by +cos x.

The next time you're not sure, draw the graph again and you'll find the sign again in front of cos for the derivative of sin. Or the sign of sin for the derivative of cos

Answer 21

f'(x)= cosx - x* sinx
f"(x)= -sinx -sinx -xcosx

Answer 22

okay got it! awesome!

Answer 23

i just couldn't understand why when i took the first der. of sin x that it gave me -cos x. I see it now though.

Answer 24

thanks again! back to coffee and filling brain cells with slopes and stuffs ha!

Answer 25

Are you still around here Ishaan94? If so I got something esle I want to ask you! haha

Answer 26

I am here!

Answer 27

Okay, haha whats the simplest way to sum up the chain rule? My book has 22 pages covering it. I am still going to work some example problems, but I don't wnat to spend three hours on it lol

Answer 28

Okay... hmm lets think of a function \(f(g(x))\)
Lets Differentiate it
\[\frac{d (f(g(x)))}{dx}*\frac{d g(x)}{dx}* \frac{dx}{dx} \]

Answer 29

I mean that is how I do it Just keep differentiating the whole function, it's like computer games clear level 1 then to level 2 clear it again then to level 3 just keep going until you reach the final level.....

Answer 30

got it

Answer 31

Let's think of some good example hmm Differentiation of \(\sin (\cos x)\)

Answer 32

Now first Differentiate sin

Answer 33

maybe you can help me with this example they give and I can figure out the rest?

this is example #1. calculate the der. of y=cos(x^3)

i have the entire solution here, I just want to follow it through right the first time. comp.functions don't scare me either. i can post this one as a new question if you want me too as well

Answer 34

Okay Let's do \(y=\cos{x^3}\)

Answer 35

lol

Answer 36

\[\frac{dy}{dx} = -\sin {x^3} \times \frac{d x^3}{dx} \]

Answer 37

\[\frac{dy}{dx} = - \sin x^3 \times 3x^2 \times \frac{dx}{dx} \]

Answer 38

\[\frac{dy}{dx} = - \sin{x^{3}} \times 3x^{2} \]

Answer 39

here check this picture out: why do they give me all this u crap under For convenience (at tope of file) that just more confusing crap in my head

Answer 40

i like to eliminate information like that, I understand the use of it, but it just gets in my way, I would just prefer to work an extra step or two then deal with that substitution stuff

Answer 41

yea just take off the substitution method,do as you want do it directly without substitution no problem with that

Answer 42

so, where did the 3 come from in the 3x^2 of the last derv. you helped me think through?

Answer 43

\(x^3\) is function of x with exponential power of 3 so when you differentiate it
\[\frac{d}{dx}x^{3} =3x^{3-1} = 3x^2\]is what you get....

Answer 44

right, i was getting ready to say that....lol

Multiply -sin(x^(3)) by 3x^(2) to get -3x^(2)sin(x^(3))
(d)/(dx) cos(x^(3))=-3x^(2)sin(x^(3))

Answer 45

thanks alot. I got about two hours left and im out of study time. lol so back to work for me. have a good day!

Answer 46

yea, good work

Answer 47

you too have a good day