Mosfet equation: how to get it?

[Granger]

Homework Statement

I'm studying the following circuit with a MOSFET

[![enter image description here][1]][1] [1]: https://i.stack.imgur.com/gEEdQ.png

Homework Equations

3. The Attempt at a Solution [/B]
Now for analyzing this circuit my book came out with various equations (which I totally understand and have no doubts about it).
Because the gate current is zero we have

$$V_G=V_{DD} \frac{R_2}{R_1+R_2}$$

Assuming the MOSFET is working on saturation we also have

$$I_D=k(V_{GS}-V_t)^2$$

And applying KVL:

$$V_{GS}=V_G-R_SI_D$$

Substituting equation (3) on equation (2) we have

$$I_D=k(V_G-R_SI_D -V_t)^2$$Now comes to the point I'm not understanding. To show us the dependence of the drain current over the threshold voltage, the book takes the derivative of the expression:

$$\frac{dI_D}{dV_t}=\frac{d}{dV_t}k(V_G-R_SI_D -V_t)^2$$

and then it writes

$$\frac{dI_D}{dV_t}=\frac{-2\sqrt{kI_D}}{1+2R_S\sqrt{kI_D}}$$

How on Earth did they go from equation (5) to equation (6) by taking the derivative. What kind of substitution are they making?

 

[NascentOxygen]

Did you try differentiating as indicated for the right hand side of your second-last equation? What do you get?

 

[Granger]

Did you try differentiating as indicated for the right hand side of your second-last equation? What do you get?

I get
$$-2k(V_G - R_SI_D-V_t)$$

 

[NascentOxygen]

Reminder: the chain rule says dy/dx =dy/dz * dz/dx

You have written the equivalent of dy/dz, and now need to multiply by the equivalent of dz/dx

 

[Granger]

Reminder: the chain rule says dy/dx =dy/dz * dz/dx

You have written the equivalent of dy/dz, and now need to multiply by the equivalent of dz/dx

What do you mean? Why using the chain rule here? I'm supposed to take the derivative of I_D in relation to V_t and that's exactly what I'm taking.

 

[NascentOxygen]

If you aren't confident in use of the chain rule, you can instead expand the squared term, i.e., get rid of the brackets on the right hand side, before attempting the differentiation. (This is a good way to check that you are going about it correctly, because obviously both methods should give identical answers.)

 

[Granger]

If you aren't confident in use of the chain rule, you can instead expand the squared term, i.e., get rid of the brackets on the right hand side, before attempting the differentiation. (This is a good way to check that you are going about it correctly, because obviously both methods should give identical answers.)

Oh I got what I was doing wrong!
I was assuming the current I_D was constant when taking the derivative which is obviously wrong. Thanks!