Question

Implicit derivative of ysin(x) = xsin(y).
I know what the answer is: (ycos(x)-sin(y))/ (xcos(y) - sin(x))
Can someone just show the steps for the implicit differentiation?

Answer 1

sinxdy/dx + ycosx = siny + xcosy dy/dx;

siny-ycosx = dy/dxsinx - xcosydy/dx;

factor out dy/dx and divide xsinx - xcosy from both sides

Answer 2

I need step by step. Like shown to me.

Answer 3

Seeing that does not really help my understanding.

Answer 4

product rule between y and sinx, product rule between x and sin(y);
when doing derivative of sinx and x you get their derivative straight away;
when doing derivative of y and sin(y) you gotta times it by dy/dx coz it's implicit diff

Answer 5

I need to see it step by step like each individual one wrote out. Like 2x+2=4
2x=4-2 --> 2x=2 --> x=2/2--> x=1. Words mean nothing, I need the actual math part elaborated to understand.

Answer 6

dy/dy * dy/dx = dy/dx; d(siny)/dy * dy/dx = d(siny)/dx, those are the concepts around why you have to times by dy/dx by the derivative of siny and y with respect to y.

Answer 7

so we start with:

\(\dfrac{d}{dx} ysin(x) \)

Answer 8

I do not understand that. Can you show me the math, please? I'm just getting more confused.

Answer 9

K

sorry i can't be of more help.

but @ikram002p is really good at explaining stuff, ..... and she's on the thread!!!

Answer 10

ysin(x) = xsin(y).

@KaleidoscopicsInk start with the right side, could you derive it with respect to x ?

Answer 11

I have no idea how to.

Answer 12

(x sin(y))'
fisrt u can use this hint
(ab)'=a'b+ab'

Answer 13

In all honesty, I just need someone to show me this problem step by step. Im tired and just need this last one done. I have no means of retaining the information. I just want this to be over with. I can do derivatives but I lack the patience to do them for a class that means absolutely nothing to me because it is honestly one of the most retarded and slow classes I have been in. If you could just do me one favor, out of the kindness of your heart and just show me how to do this problem, i'd be forever grateful.

Answer 14

so
(x sin y)'=x'(sin y)+x (sin y')
x'=1
(sin y)' you can do this by the implicit differentiation rule which is
[f(g(x))]'=f' * g'
thus
(sin y)'=cos y* y'

y'=dy/dx right ?

Answer 15

Sure. I really don't know. I just want this over with, please.

Answer 16

well, i don't think your interesting in learning this. and i'm sure i'm not interested in teaching this to people who keep whining and not listening. i'm trying to do step by step solution to u and u are like (please i wanna this over).
hold your self up and try to focus WOULD YOU!

Answer 17

As hard as it is to believe, I am. I have been. I'm trying not to break at the moment, so sittting here and being confused and nearly lectured is difficult to bare because all I want is this class to be over and my head is killing me. I'm driving myself insane with this one question. I'm trying, if you believe it or not, that's your choice. But I'm freaking out and my calm is completely gone. I'm sorry. I am if I come off that way. Im not trying to be rude, I'm trying not to break.

Answer 18

just keep calm and take a rest =). it doesn't worth all that stress for real.


back to the topic
now " implicit differentiation " is used when you two or more variables sometimes they are not separable.

Answer 19

I know, I can't rest. I get yelled at if I do.

Answer 20

I know what implicit differentiation is.

Answer 21

you know what it is. good !

what is the derivative of sin(x) ?

Answer 22

cos(x)

Answer 23

good!
and what is derivative of sin(y) with respect to x (dy/dx) ?

Answer 24

I don't know.

Answer 25

use this rule

f(y)'=f'(y)*y'

Answer 26

give a try if you couldn't find it i'll tell you how.

Answer 27

cos(x)sin(y)

Answer 28

very close
but y'=dy/dx

Answer 29

I don't know.

Answer 30

well
sin(y)'=cos y*y'=cos y *(dy/dx)

Answer 31

And?

Answer 32

\(\Large

{\color{Red} { \text{remember } }} (ab)'=ab'+a'b
\\
\begin{matrix}
(y~\sin(x))' = (x ~\sin(y) )' \\
y(\sin (x))'+y'\sin(x)=x(\sin(y))'+x'(\sin(y)) \\
y \cos(x)+\dfrac{dy}{dx} \sin (x) =x \cos(y)\dfrac{dy}{dx} + \sin (y) \\

\end{matrix} \)

Answer 33

Okay, I know the product rule.

Answer 34

seems u know everything now cool !

Answer 35

Can you show me how to do the problem now?

Answer 36

well check the last line we have concluded lately
\( \large y \cos(x)+{\color{Red} {\dfrac{dy}{dx}}} \sin (x) =x \cos(y){\color{Red} {\dfrac{dy}{dx}}} + \sin (y)
\)

Answer 37

all you need now is to put dy/dx on a side on it's own could you do that ?

Answer 38

No.

Answer 39

assume you have this problem

\(\Large a+ b\times c = d\times c +e\) can you solve for c ?(put c on a side on it's own?)

Answer 40

I cannot. I'm getting yelled at. My calm is lost.

Answer 41

-,-

Answer 42

Not yelled at by you. Other people.

Answer 43

ok here is away...
collect all terms with c on the same side
bc-dc=e-a

make c a common factor
c(b-d)=e-a
let c alone
c=(e-a)/(b-d)
simple algebra !

Answer 44

now if i ask you would you do that for dx/dy you would complain and not giving a try

Answer 45

I don't understand. I really don't.

Answer 46

ok here is another question

3x+4=x+2

solve for x

Answer 47

I cannot do that with the given trig things. Im bad at it.

Answer 48

\(\large
\begin{matrix}
y \cos(x)+{\color{Red} {\dfrac{dy}{dx}}} \sin (x) =x \cos(y){\color{Red} {\dfrac{dy}{dx}}} + \sin (y) \\
{\color{Red} {\dfrac{dy}{dx}}} \sin (x)-x \cos(y){\color{Red} {\dfrac{dy}{dx}}} = \sin (y) -y \cos(x)\\
{\color{Red} {\dfrac{dy}{dx}}} [\sin (x)-x \cos(y)] = \sin (y) -y \cos(x) \\
{\color{Red} {\dfrac{dy}{dx}}} = \dfrac {\sin (y) -y \cos(x)}{ \sin (x)-x \cos(y)} \\

\end{matrix} \)

Answer 49

Thank you so much for clarifying that!!!! My gratitude is beyond belief.

Answer 50

full solution would be like:-
\( \Large
\begin{matrix}
(y~\sin(x))' = (x ~\sin(y) )' \\
y(\sin (x))'+y'\sin(x)=x(\sin(y))'+x'(\sin(y)) \\
y \cos(x)+\dfrac{dy}{dx} \sin (x) =x \cos(y)\dfrac{dy}{dx} + \sin (y) \\


y \cos(x)+{\color{Red} {\dfrac{dy}{dx}}} \sin (x) =x \cos(y){\color{Red} {\dfrac{dy}{dx}}} + \sin (y) \\
{\color{Red} {\dfrac{dy}{dx}}} \sin (x)-x \cos(y){\color{Red} {\dfrac{dy}{dx}}} = \sin (y) -y \cos(x)\\
{\color{Red} {\dfrac{dy}{dx}}} [\sin (x)-x \cos(y)] = \sin (y) -y \cos(x) \\
{\color{Red} {\dfrac{dy}{dx}}} = \dfrac {\sin (y) -y \cos(x)}{ \sin (x)-x \cos(y)} \\

\end{matrix}
\)

Answer 51

PS:-
\( \dfrac { y \cos(x)-\sin (y) }{ x \cos(y)-\sin (x)} = \dfrac {\sin (y) -y \cos(x)}{ \sin (x)-x \cos(y)}
\)