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Mathematics 10 Online
OpenStudy (anonymous):

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.

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

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?

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

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

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

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

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

OpenStudy (anonymous):

Yeah I did all that and got y=-x+1

OpenStudy (anonymous):

yeah, looks good

OpenStudy (anonymous):

Okay cool, thanks man

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