Question

If tan(xy)=x , thendy/dx=?

Answer 1

You need to find dy/dx, right? So the first thing you need to do is to isolate y.
Hence, use inverse trig function for that.
xy = arctan(x)

Answer 2

Then
y = (arctanx)/x
Are you with me so, far?

Answer 3

Yes

Answer 4

Now, do quotient rule. Can you?

Answer 5

Is it y'= ((x) (1/(x^2+1)) - arctanx ) / x^2 ?

Answer 6

You're on the right track. Just simplify it.

Answer 7

Can I get some help with that? :/

Answer 8

\[y' =\frac{ \frac{ x }{ x^2+1 } - \tan^{-1} x }{ x^2 }\]
Hence
\[y' = \frac{ x- \tan^{-1} x (x^2+1)}{ x^2(x^2+1) }\]

Answer 9

Ok I understand how you did that. But now I don't know which answer choice that one is. May I show you the choices?

Answer 10

Sure.

Answer 11

A) 1- ytan(xy)sec(xy) / xtan(xy)sec(xy
B) sec(xy)^2-y / x
C) cos(xy)^2
D) cos(xy)^2 / x
E) cos(xy)^2 - y / x

Answer 12

Sure, you should just do implicit differentiation. All that means is every time you see y, you do the chain rule. I suggest you ALWAYS pretend you're doing implicit differentiation. For instance:

\[y=x^2\]
\[\frac{ d }{ dx }(y)=\frac{ d }{ dx }(x^2)=2x\]
Now we have always done it this way, but with y by itself we sort of didn't give it much thought on the left hand side. But you must stop and think about what if you just took the derivative of y by itself with respect to x without knowing what y is equal to?

\[\frac{ d }{ dx }(y)=\frac{ dy }{ dx }\]

So that's really just saying, we don't know. The derivative of y is equal to dy/dx. Not surprising, but now when we rearrange that example I'm showing you, we leave this idea of dy/dx as just simply being an unknown in there:
\[y=x^2\]
Start back there, square root both sides (remember, same as to the 1/2 power)
\[y^{1/2}=x\]

Now what happens when we take the derivative?
\[\frac{ d }{ dx } (y^{1/2})=\frac{ d }{ dx }(x)\]

See the left side we must do the chain rule, and the right side is just simply 1 like normal.
\[\frac{ 1 }{ 2 }y^{(-1/2)}*\frac{ dy }{ dx }=1\]

This is crucial to understanding. Now at first this might look like a completely different problem and answer than y=x^2. BUT IT IS EXACTLY THE SAME. A little bit of algebra by solving for the dy/dx term on the left side and plugging in y to the this derivative from the original equation, y=x^2 will give you the exact same answer,

dy/dx=2x.

Even though this is a longer, different way, the point is to illustrate that it is completely consistent whether you can solve for y or not before taking the derivative.

So if you understand this, apply it to the problem you've been given and I will help you get the right answer. =)

Answer 13

So am I doing this right?
d/dx (tan(xy)) = d/dx (x)
sec(xy)^2 (xy'+y) = 1

Answer 14

Yes exactly right. Now just solve for y' and remember than 1/sec is cos

Answer 15

So y' = ( (1/(sec(xy))^2 ) - y ) / x ?

Answer 16

Which appears to be the same as E.

Answer 17

Ohhh okay. Thank you very much!