Find an equation of the tangent line to the graph of the equation at the given point.
arctan (x+y)=y^2 + pi/4 point: (1,0)
Work so far attached, I'm stuck.
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OpenStudy (anonymous):
OpenStudy (anonymous):
are you working in this?
OpenStudy (anonymous):
You mean the OS Equation Editor? No
OpenStudy (anonymous):
i meant the first equation
OpenStudy (anonymous):
I have to go to bed so feel free to input anything that will help me solve the problem and I'll check this question in the morning
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OpenStudy (anonymous):
i think it is fast
OpenStudy (anonymous):
well yes, my work is attached
OpenStudy (anonymous):
can you wait some minutes?
OpenStudy (anonymous):
Okay, sure
OpenStudy (anonymous):
only the first equation , right?
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OpenStudy (anonymous):
there's only one equation, so yes
OpenStudy (anonymous):
Yes, the formula is u'/(1+u)^2
OpenStudy (anonymous):
Yes... I did that but what confused me was that I had y' on both sides
OpenStudy (anonymous):
wait
OpenStudy (anonymous):
y' is the slope of the tangent
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OpenStudy (anonymous):
at point (1,0)
OpenStudy (anonymous):
it means to evaluate it for x=1 and y=0
OpenStudy (anonymous):
but you need y' on one side to get the slope of the tangent, right?
OpenStudy (anonymous):
wait me 3 minuts please
OpenStudy (anonymous):
plug in the numbers then solve it is easier
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OpenStudy (anonymous):
But I need to find the actual derivative before I can solve and that's what's tripping me up
OpenStudy (anonymous):
oh i see ok
\[\tan^{-1}(x+y)=y^2\] to start, then
\[\frac{y'+1}{(x+y)^2+1}=2yy'\]
OpenStudy (anonymous):
right so since the y' is on both sides what do I do
OpenStudy (anonymous):
factor it out right
OpenStudy (anonymous):
replace it for the point that you want
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OpenStudy (anonymous):
you have a choice
ugly algebra, or plug in the numbers now
OpenStudy (anonymous):
Okay... so numbers
OpenStudy (anonymous):
i would choose the second one because it is much much easier
OpenStudy (anonymous):
so what I just plug in for x and y?
OpenStudy (anonymous):
replace \(x\) by 1, \(y\) by 0
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OpenStudy (anonymous):
point: (1,0)
OpenStudy (anonymous):
so then ?
OpenStudy (anonymous):
\(\frac{y'+1}{2}=0\) you can just about solve for \(y'\) in your head
OpenStudy (anonymous):
so my work is correct?
OpenStudy (anonymous):
slope is \(-1\) right? then i guess point - slope to get the line
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