Simple Implicit Differenentiation Help

[Asphyxiated]

Homework Statement

The original function is given as such:

$$y = 13 arctan(\sqrt{x})$$

Homework Equations

The Attempt at a Solution

I went ahead and changed it into:

$$tan(\frac{y}{13}) = \sqrt{x}$$

I thought it would be simpler this way. So now I differentiate:

$$\frac{1}{13} sec^{2}(\frac{y}{13})\frac{dy}{dx} = \frac {1}{2\sqrt{x}}$$

Notice that the 1/13 came from the y/13 derivative, as obviously this is why dy/dx is there as well.

So it should be this:

$$\frac {dy}{dx} = \frac{\frac{1}{2\sqrt{x}}}{\frac{sec^{2}(\frac{y}{13})}{13}}$$

which is:

$$\frac {dy}{dx} = \frac {1}{2\sqrt{x}} * \frac {13}{sec^{2}(\frac{y}{13})}$$

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(sec^{2}(\frac{y}{13}))}$$


Where did I go wrong? This is evidently not the correct answer so... yeah. I have done it over a couple times but perhaps my algebra in the very beginning doesn't hold true, I don't know.

Thanks!

 

[Dick]

Homework Statement

The original function is given as such:

$$y = 13 arctan(\sqrt{x})$$

Homework Equations

The Attempt at a Solution

I went ahead and changed it into:

$$tan(\frac{y}{13}) = \sqrt{x}$$

I thought it would be simpler this way. So now I differentiate:

$$\frac{1}{13} sec^{2}(\frac{y}{13})\frac{dy}{dx} = \frac {1}{2\sqrt{x}}$$

Notice that the 1/13 came from the y/13 derivative, as obviously this is why dy/dx is there as well.

So it should be this:

$$\frac {dy}{dx} = \frac{\frac{1}{2\sqrt{x}}}{\frac{sec^{2}(\frac{y}{13})}{13}}$$

which is:

$$\frac {dy}{dx} = \frac {1}{2\sqrt{x}} * \frac {13}{sec^{2}(\frac{y}{13})}$$

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(sec^{2}(\frac{y}{13}))}$$


Where did I go wrong? This is evidently not the correct answer so... yeah. I have done it over a couple times but perhaps my algebra in the very beginning doesn't hold true, I don't know.

Thanks!

That's perfectly correct. There's a simpler way to write the answer. Use that sec(y/13)^2=1+tan(y/13)^2. You could have gotten there directly by just differentiating the arctan.

 

[Asphyxiated]

Ok so basically you are saying that:

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(sec^{2}(\frac{y}{13}))}$$

is:

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(1+tan^{2}(\frac{y}{13}))}$$

so then I get:

$$\frac {dy}{dx} = \frac {13}{2\sqrt{x}+2\sqrt{x}tan^{2}(\frac{y}{13})}$$

right? I tried this answer as well and it is evidently not correct either.

The question does say to simplify as much as possible but unless I need to make more trig subs I don't see how that is possible.

thanks again!

 

[Dick]

Ok so basically you are saying that:

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(sec^{2}(\frac{y}{13}))}$$

is:

$$\frac {dy}{dx} = \frac {13}{(2\sqrt{x})(1+tan^{2}(\frac{y}{13}))}$$

so then I get:

$$\frac {dy}{dx} = \frac {13}{2\sqrt{x}+2\sqrt{x}tan^{2}(\frac{y}{13})}$$

right? I tried this answer as well and it is evidently not correct either.

The question does say to simplify as much as possible but unless I need to make more trig subs I don't see how that is possible.

thanks again!

tan(y/13)=sqrt(x). That's what you said in the first post. Use that to simplify.

 

[Asphyxiated]

Yeah I got it after you posted! Thanks though, quite a bit.