Implicit Differentiation with Trig

1. Oct 20, 2007

bondgirl007

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Oct 20, 2007

rocomath

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?

Last edited: Oct 20, 2007
3. Oct 20, 2007

bondgirl007

I derived it and I got:

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

Is that the right derivative?

4. Oct 20, 2007

rocomath

yeppp

5. Oct 20, 2007

bondgirl007

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. Oct 20, 2007

rocomath

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. Oct 20, 2007

bondgirl007

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. Oct 20, 2007

rocomath

$$x\tan{y}=y-1$$ @ $$(x,\frac{\pi}{4})$$

Last edited: Oct 20, 2007
9. Oct 20, 2007

bondgirl007

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. Oct 20, 2007

rocomath

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. Oct 20, 2007

bondgirl007

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?

Last edited: Oct 20, 2007
12. Oct 20, 2007

rocomath

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

Last edited: Oct 20, 2007
13. Oct 21, 2007

bondgirl007

I got the same as numerator as you for the derivative but my denominator is xsec^2 y - 1

14. Oct 21, 2007

rocomath

it's equivalent

15. Oct 21, 2007

ali1982

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. Oct 21, 2007

rocomath

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

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

check plz!

Last edited: Oct 21, 2007
17. Oct 21, 2007

coomast

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. Oct 21, 2007

ali1982

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.