Implicit Differentiation with Trig

  • #1

Homework Statement



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


Homework Equations





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.
 

Answers and Replies

  • #2
1,752
1
you miswrote it

[tex]x\sec^{2}y\frac{dy}{dx}+tan{y}=\frac{dy}{dx}[/tex]

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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  • #3
I derived it and I got:

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

Is that the right derivative?
 
  • #4
1,752
1
I derived it and I got:

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

Is that the right derivative?
yeppp
 
  • #5
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.
 
  • #6
1,752
1
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?
 
  • #7
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.
 
  • #8
1,752
1
[tex]x\tan{y}=y-1[/tex] @ [tex](x,\frac{\pi}{4})[/tex]
 
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  • #9
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
1,752
1
that's not the X value i got, check it again

[tex]x=\frac{y-1}{\tan{y}}[/tex]

the derivative is the slope of your tangent line at your (x,y) so once you find both of those, you can find [tex]\frac{dy}{dx}[/tex]
 
  • #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
1,752
1
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. [tex]\frac{\pi}{4}=1[/tex]

i make too many mistakes!

[tex]y'=\frac{dy}{dx}=\frac{\tan{y}}{1-x\sec^{2}y}[/tex] @ (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
1,752
1
I derived it and I got:

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

Is that the right derivative?
it's equivalent
 
  • #15
2
0
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
1,752
1
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
[tex][x\tan{y}=y-1] \neq [\frac{dy}{dx}=\sec^{2}y\frac{dy}{dx}+\tan{y}][/tex]

[tex][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}][/tex]

check plz!
 
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  • #17
279
0
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
2
0
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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