Implicit Differentiation with Trig

bondgirl007
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Homework Statement



Find the slope of the tangent line to x tan y = y - 1 when y = pi/4


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The Attempt at a Solution



I can't seem to get the derivative. Here's what I do.

First I used the product rule the left side of the equation and got sec^2 x dy/dx + tan y = dy/dx.

When I simplify, I get dy/dy over dy/dx, which just reduces to 1 so I'm having trouble with getting dy/dx on it own.
 
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you miswrote it

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

but either way you should have been able to algebraically solve for dy\dx

what would your next step be to find x?
 
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I derived it and I got:

dy/dx = -tany/(xsec^2 - 1).

Is that the right derivative?
 
bondgirl007 said:
I derived it and I got:

dy/dx = -tany/(xsec^2 - 1).

Is that the right derivative?
yeppp
 
Thanks for the help, people!

I need help with the tangent line now though. I have to find it when y = pi/4.
I'm not sure how to approach it because I don't have the value for x.
 
you can easily find it, you're given your y value. you also have your original equation, correct? where does the tangent line pass through?
 
It passes through pi/4.

Here's my equation:

dy/dx = -tan(pi/4)/(xsec^2 - 1)

I have two unknowns so not sure how to find it.
 
x\tan{y}=y-1 @ (x,\frac{\pi}{4})
 
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For x, I got:

pi/4 - 1

When I find the slope and plug these in, I don't get what to do with the x and sec^2. Is x part of the sec or not?
 
  • #10
that's not the X value i got, check it again

x=\frac{y-1}{\tan{y}}

the derivative is the slope of your tangent line at your (x,y) so once you find both of those, you can find \frac{dy}{dx}
 
  • #11
Thanks for the reply. I solved for x and still got: pi/4 - 1.

I'm not sure how to get the derivative by subbing in for dy/dx. What should I do with the sec^2. Is x part of the sec^2 function or is it a coefficient?
 
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  • #12
bondgirl007 said:
Thanks for the reply. I solved for x and still got: pi/4 - 1.

I'm not sure how to get the derivative by subbing in for dy/dx. What should I do with the sec^2. Is x part of the sec^2 function or is it a coefficient?
lol I'm so sorry, had a dumb moment. \frac{\pi}{4}=1

i make too many mistakes!

y'=\frac{dy}{dx}=\frac{\tan{y}}{1-x\sec^{2}y} @ (x,y) = m = slope
 
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  • #13
I got the same as numerator as you for the derivative but my denominator is xsec^2 y - 1
 
  • #14
bondgirl007 said:
I derived it and I got:

dy/dx = -tany/(xsec^2 - 1).

Is that the right derivative?

it's equivalent
 
  • #15
This is the reply for bond girl007

Solution:
xtany=y-1
differentiate the two sides of the equation:
d/dx(xtany) =d/dx(y-1) Lows of differentiation

tany+sec^2(y)dy/dx=dy/dx

make dy/dx in one side: tany=(1-sec^2y)dy/dx

using laws of trigonometrics: tany=(-tan^2y)dy/dx

we get: 1=-tanydy/dx

dy/dx=-1/tany

dy/dx=-1/tan(pi/4)

Hence: dy/dx=-1
 
  • #16
ali1982 said:
Solution:
xtany=y-1
differentiate the two sides of the equation:
d/dx(xtany) =d/dx(y-1) Lows of differentiation

tany+sec^2(y)dy/dx=dy/dx

make dy/dx in one side: tany=(1-sec^2y)dy/dx

using laws of trigonometrics: tany=(-tan^2y)dy/dx

we get: 1=-tanydy/dx

dy/dx=-1/tany

dy/dx=-1/tan(pi/4)

Hence: dy/dx=-1
[x\tan{y}=y-1] \neq [\frac{dy}{dx}=\sec^{2}y\frac{dy}{dx}+\tan{y}]

[x\tan{y}=y-1] \mbox{should equal, if I'm not mistaken b/c it's almost 2am} [x\sec^{2}y\frac{dy}{dx}+\tan{y}=\frac{dy}{dx}]

check please!
 
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  • #17
Indeed, ali1982 forgot the x in his equation. The result x=pi/4-1 is indeed correct. Now it is not difficult to find the slope. It is allready derived as dy/dx=tan(y)/[1-x*sec^2(y)]. The numbers x and y have been obtained, just put them in here. I think yo should get something like 2/(6-pi) approx. 0.7
 
  • #18
thanYes, I forgot the x.When we differentiate, the term : sec^2ydy/dx must be multiplied by x ,and this term will be : xsec^2ydy/dx.
Thanks for correction.
 
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