Question

I will risk continuing this without shamil98.

Answer 1

I learned so far the rule of power and rule of product doing derivatives.

Can someone esle continue tutoring me?

Answer 2

Have you learnt the chain rule?

Answer 3

No

Answer 4

The chain rule:
\[\frac {dy}{dx}=\frac {dy}{du}\cdot\frac{du}{dx}\]

Answer 5

For example, y=tan(x^2).

Try to find \(\dfrac{dy}{dx}\) using the above formula with \(\large u=x^2\) :)

Answer 6

I never learned this in a dx/dy form, I guess the risk turned out a bad way.

Answer 7

oh, what is the notation that you learnt?

Answer 8

I'll try to convert the formula to your notation

Answer 9

\[nx^{x-1}\] like \[x^4~'=4x^3\]

AND \[f'g+g'f\]

Answer 10

I guess shamil98 would be my best option....

Answer 11

\[y'(x)=y'(u)\times u'(x)\]
y(x) means y as a function of x.

Answer 12

what does u represent?

Answer 13

A function of x, for example,

Answer 14

if I want to find the derivative of -ln(cos(x)):

Answer 15

YEah, I am barely able to find regular derivatives, hold on.....

Answer 16

Let u=cos(x).
-ln(cos(x))'
= -ln(u)' * u'
= -1/u * -sin(x)
= -1/cos(x) * -sin(x)
= tan(x)

Answer 17

I can';t find the previous q, b.c my prof isn't working.

Answer 18

This is what I was left off.
http://openstudy.com/users/shamil98#/updates/52b94eb1e4b01cdca4494f8a

Answer 19

ok, do you know the derivatives of the trigonometric functions?

Answer 20

I've never done that

Answer 21

ok, let me teach you two basic derivatives of trigonometric functions:
\[\sin(x)'=\cos(x)\]\[\cos(x)'=-\sin(x)\]

Answer 22

Let's practice what you learnt :)
Try to find the derivative of \(\tan(x)\) using the quotient rule :)

Answer 23

A;right, I was about to say nvm....
I need to know the definitions first using easier examples, then do harder and harder. If you go through that question, you will see that I am incapable of deriving tan(x)

Answer 24

But @shamil98 just taught you the quotient rule...

Answer 25

Got disconnected.... but not how to apply it to trigonometric function though.

Answer 26

Yeah, I'll wait for shamil. kennylau, ty for your help though

Answer 27

no :/
I can help you :/

Answer 28

Sure?

Answer 29

ok, find the derivative of \(\dfrac{x^2+1}{x+3}\) using the quotient rule :)

Answer 30

\[2x(x+3)+1(x^2+1)\] right?

Answer 31

\[f'g+g'f\]

Answer 32

That is the product rule :)

Answer 33

The which one is the quotient?

Answer 34

\[(f/g)'=(f'g-g'f)/g^2\]

Answer 35

Oh, that part is the unclear part, I stopped like right before that, or at least meant to.

Answer 36

So let me give you the formulae:
\[\begin{array}{rcl}(f+g)'&=&f'+g'\\(f-g)'&=&f'-g'\\(fg)'&=&f'g+g'f\\\left(\dfrac fg\right)'&=&\frac{f'g-g'f}{g^2}\end{array}\]

Answer 37

as well as:
\[\Large(x^n)'=nx^{n-1}\]

Answer 38

Yeah, the last think I do know, it's the power rule. and I get the rest too.

Answer 39

So do you still wanna do my question? :)

Answer 40

Hold on, I need to get some staff down.

Answer 41

ok

Answer 42

Continue to wait for @shamil98 coz I'm in GMT+8 and I need to sleep now :)

Answer 43

Pk, I'll continue later, I need rest.

Answer 44

What is your timezone? :)

Answer 45

11:13 AM US, eastern coast.