implicit differentiation

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whiteti

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Use implicit differentiation to find y' from xy=sec(x+y)

this is not like any of the examples given in class so I am really stuck on where to begin. Please help :)
 
whiteti said:
Use implicit differentiation to find y' from xy=sec(x+y)

this is not like any of the examples given in class so I am really stuck on where to begin. Please help :)
Click to expand...

Use product rule on the left and chain rule on the right.
 
so to start it should look like
(x)'(y)+(x)(y)'= (secx)'*(secy)' ?

Now what?
 
whiteti said:
so to start it should look like
(x)'(y)+(x)(y)'= (secx)'*(secy)' This is incorrect

Now what?
Click to expand...

d/dx[sec(x+y)] = sec(x+y)*tan(x+y) * [1 + y']
 
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ok, I see where I went wrong there.

So now I have to isolate y' right?
how do I do that?
 
No, that's not even close.

Subhotosh Kahn told you that d/dx[sec(x+y)] = sec(x+y)*tan(x+y) * [1 + y']
Do you not see that there will be a term with y' multiplied by sec(x+y)tan(x+y)?
And so when you solve for y', there will be a term involving sec(x+y)tan(x+y) in the denominator?
 
HallsofIvy said:
No, that's not even close.

Subhotosh Kahn told you that d/dx[sec(x+y)] = sec(x+y)*tan(x+y) * [1 + y']
Do you not see that there will be a term with y' multiplied by sec(x+y)tan(x+y)?
And so when you solve for y', there will be a term involving sec(x+y)tan(x+y) in the denominator?
Click to expand...

Yeah, I totally realized what I did wrong. Sorry, just a little algebra mistake.
It'd be y'=sec(x+y)*tan(x+y)-y / x-sec(x+y)-tan(x+y)
 
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