Question

Help! I want to know what implicit derivate is the following:

tan(y) =xy

Please, I will give you medals :)

Answer 1

think of it as \[\tan(f(x))=xf(x)\]

Answer 2

chain rule on the left, product rule on the right

Answer 3

thanks my friend, but I would like to know the result to see if your results matches with mine

Answer 4

if that is not clear, try this one \[\tan(\sin(x))=x\sin(x)\] and take derivatives there

Answer 5

if you are just looking for an answer, type it in to wolfram
it will give you \(y'(x)\) and also \(x'(y)\)

Answer 6

@agent0smith @zepdrix

Answer 7

Glad to hear you've already done the problem. Please share your work. That way I'd have something to comment on.

Answer 8

I need the result @mathmale to see if yours matches with mine

Answer 9

Some people might share their answers with you. My personal preference is that you share your work with me first.

Answer 10

2sec(y)*sec(y)*tan(y)*dy/dx =dy/dx

Answer 11

dy/dx not correct, because you have a PRODUCT on the right side: xy.

You must use the product and chain rules to differentiate that.

Answer 12

I'm speaking of your result dy/dx on the right side of your equation.

Answer 13

The derivative of xy, with respect to x, comes out as (x)(dy/dx) + ( )( )

Answer 14

I already got the answer because my math guide gives me the result... I just want to know if your results matches with mine... I am not asking for the procedure

Answer 15

Sorry, but as I explained before, my pref is to see your work before I share mine.

You have xy on the right side of your original equation; your attempt to differentiate this product xy did not succeed. You need to use the product and chain rules to differentiate xy with respect to x.

Answer 16

More important than the answer is HOW you correctly arrive at it. That's why I'm being so insistent on your showing me your work.
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Answer 17

Differentiating tan(y) =xy requires that you know the derivative of the tangent function and recognize that y is itself a separate function. Thus, after using the rule for the derivative of the tangent function, you must apply the chain rule: find the derivative of y with respect to x.

Answer 18

Your "2sec(y)*sec(y)*tan(y)*dy/dx" is partially right in that you ended it with (dy/dx), which is the deriv. of the function y with respect to x. As for the earlier part, please review the derivatives of trig functions, specifically the derivative of the tangent function.

Answer 19

to be honest, I know how to solve the exercise, but a friend of mine is sitting next to me and is seeing this webside for fisrt time. He asks if you are able to solve the exercise, because he did it... actually, he is impressed by this webside

Answer 20

I'm so glad he's impressed.

I just do not understand your reluctance (or your friend's) to share all of your own work. You have already shared some work, and I have given you specific feedback on it, on both sides of the equation containing your result.

What is the derivative of tan x?

Answer 21

If it's your friend who wants to know how to differentiate implicitly, put him on; no need for you to act as middleman.

Answer 22

\[\frac{ d }{ dx }\left\{ \tan \left( y \right) \right\}=\sec ^2y*\frac{ dy }{ dx }\]

Answer 23

sshayer: correct. But I wanted crisulcampo to find that.

Answer 24

\[\sec^2(y)*(dy/dx) = (dx/dx)*y + (dy/dx)*x\]
\[\sec^2(y)*(dy/dx) -(dy/dx)*x = y\]
\[(dy/dx)[\sec^2(y)-x] = y\]
\[dy/dx= y/[\sec^2(y) - x]\]
\[dy/dx = \frac{ y }{ \frac{ 1 }{ \cos^2(y) }-x }\]
\[dy/dx= \frac{ y }{ \frac{ 1-xcos^2(y) }{ \cos^2(y) } }\]
\[dy/dx= \frac{ ycos^2(y) }{ 1-xcos^2(y) }\]

Answer 25

thanks for going thru the trouble to type all that out. Was that you, or was that your friend?

I'm not saying whether your result is right or wrong. I'll tell you, however, that the method generally used is MUCH shorter:\[\tan y = xy\]

Answer 26

Differentiating the left side with respect to x,\[\sec^2 y \frac{ dy }{ dx }\]

Answer 27

that employed both the derivative rule for the tangent function and also the chain rule.

Differentiating the right side (xy, a product), with respect to x gives us\[x \frac{ dy }{ dx }+y \frac{ dx }{ dx }\]

Answer 28

which simplifies to x(dy/dx) + y. You now have to put that equation back together and solve for (dy/dx). Do that and you'll have your answer, the derivative dy/dx.

Answer 29

\[\sec^2 y \frac{ dy }{ dx }=x \frac{ dy }{ dx}+y\]

Answer 30

How would you solve that for (dy/dx)?

Answer 31

Hint: move all the (dy/dx) terms to the left side. This leaves you with y alone on the right side.

Answer 32

Factor (dy/dx) out of the left side. How would you get rid of the other factor of the left side? You want to end up with (dy/dx) alone.

Answer 33

my process is right my friend.... I know dude but my friend wanted to know that you were able to solve the exercise

Answer 34

the one who did the proccess shown was me

Answer 35

My friend only wanted to know if you were able to get my answer

Answer 36

Why would your friend want to know whether I, a complete stranger, were bright enough to solve this problem? I'm not very pretty to look at.

Answer 37

As I told you before,he is knowing this webside for first time, so I wanted him to realize that most clever community that belongs to this webside is able to solve an exercise of his own desicion

Answer 38

that's all

Answer 39

but you made it complicated

Answer 40

Sorry you found it that way. I always try to get others to do as much as they can on a given problem and to share work with me before I obtain my own answers and share them. You were equally stubborn about sharing your own work. Draw?

Encourage your friend to sign up for OpenStudy. But be aware that OpenStudy is closing at the end of January. I'm very sad about that.

Answer 41

May I pose a challenge to you two?

Show that your \[dy/dx= \frac{ ycos^2(y) }{ 1-xcos^2(y) }\]

Answer 42

is the same result as I would have gotten had I completely finished finding (dy/dx).