Implicit Differentiation with Trig

[bondgirl007]

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.

 

[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?

 

[bondgirl007]

I derived it and I got:

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

Is that the right derivative?

 

[rocomath]

I derived it and I got:

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

Is that the right derivative?

yeppp

 

[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.

 

[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?

 

[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.

 

[rocomath]

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

 

[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?

 

[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 $$\frac{dy}{dx}$$

 

[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?

 

[rocomath]

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

 

[bondgirl007]

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

 

[rocomath]

I derived it and I got:

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

Is that the right derivative?

it's equivalent

 

[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

 

[rocomath]

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!

 

[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

 

[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.