Using dx/dy to find a y-parallel tangent

[Faiq]

Homework Statement

The curve $C$ has equation
$$y=x^2+0.2sin(x+y)$$
Show that $C$ has no tangent(no point where $dy/dx=∞$), that is parallel to the y axis.

Attempt
$$1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$$

For a tangent to be parallel to y-axis,

$$\frac{dx}{dy}=0$$

$$1=0.2cos(x+y)$$

$$cos(x+y)=5$$

No value of $(x,y)$ exist for which $cos(x+y)>1$, hence no y-axis parallel tangent.

**Confusion**
Never been taught or encountered $\frac{dx}{dy}$ and thus I am having doubts over the validity of the solution.

Can somebody please verify it?

 

[BvU]

What about $dx\over dy$ if ${dy\over dx} < Z$ where you can show that Z is finite ?

 

[Mark44]

Thread moved. @Faiq, in the future, please post questions about derivatives in the Calculus & Beyond section, not the Precalc section.

 

[Mark44]

Why not just find dy/dx, and show that it is defined for all real x?

 

[BvU]

What about $dx\over dy$ if ${dy\over dx} < Z$ where you can show that Z is finite ?

OK, ' is finite for all finite x ' -- thanks, Mark !

 

[Faiq]

What about $dx\over dy$ if ${dy\over dx} < Z$ where you can show that Z is finite ?

Not sure what you're telling. Can you please elaborate?

 

[Faiq]

Why not just find dy/dx, and show that it is defined for all real x?

I am just concerned whether my solution is correct or not.

 

[Mark44]

I am just concerned whether my solution is correct or not.

Your solution looks fine to me.

 

[Buffu]

Homework Statement

The curve $C$ has equation
$$y=x^2+0.2sin(x+y)$$
Show that $C$ has no tangent(no point where $dy/dx=∞$), that is parallel to the y axis.

Attempt
$$1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$$

For a tangent to be parallel to y-axis,

$$\frac{dx}{dy}=0$$

$$1=0.2cos(x+y)$$

$$cos(x+y)=5$$

No value of $(x,y)$ exist for which $cos(x+y)>1$, hence no y-axis parallel tangent.

**Confusion**
Never been taught or encountered $\frac{dx}{dy}$ and thus I am having doubts over the validity of the solution.

Can somebody please verify it?

I guess it is correct because by inverse function theorem $D_x (y) = 1/D_y(x)$. So if $D_y(x) \to \infty$, $D_x(y) \to 0$.

 

[SammyS]

I am just concerned whether my solution is correct or not.

In that case, you may need to show how it is that requiring dx/dy = 0 gets you from

$\displaystyle 1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$​

to

$\displaystyle 1=0.2cos(x+y)$​

.

 

[Faiq]

Okay thanks to all of you very much.