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If one has to use the chain rule repeatedly, such as in the problem $y=\cos(2u)$ and $u=3x+1$, would $u$ be viewed as a constant? There's an easier way to solve this function by simply making $u=6x+2$ but this is just for an example.

I know that the derivative $\frac{du}{dx}$ is $3$.

I know that the derivative of $\cos(2u)$ would be equal to $-2\sin(2u)$ since $2u=j$ and $f'(\cos(j))\times f'(2u)=-2\sin(2u)$. However, in solving $\cos(2u)$, its actual value of $3x+1$ was removed, since the actual derivative of $2u$ with the value of $u$ plugged in would be $6$.

Why would you treat $u$ as completely independent from its actual value when using the chain rule (until you plug it back in)?

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Nothing is being removed. It’s just another way of reaching the same answer. According to the Chain Rule, we have the following:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ when $y = f(u)$ and $u = g(x)$

Here, $u = 3x+1$ and $y = \cos(2u) \implies y = \cos(2(3x+1))$.

$$\frac{dy}{dx} = \frac{d}{du}\cos(2u) \cdot \frac{d}{dx}(2u)$$

$$\frac{dy}{dx} = (2u)’\cdot-\sin(2u)$$

$$u = 3x+1 \implies \frac{dy}{dx} = (6x+2)’\cdot -\sin(6x+2)$$

$$\frac{dy}{dx} = -6\sin(6x+2)$$

In examples like this, of course, solving for $y = \cos(6x+2)$ is pretty quick, getting the same answer, just as you mentioned.

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In this case we can proceed by chain rule

$$\frac{dy}{dx}=\frac{d \cos(2u)}{du}\frac{du}{dx}$$

or by a direct calculation by substitution

$$\frac{dy}{dx}=\frac{d \cos(2(3x+1))}{dx}$$

to derive and nothing is keep constant here.

We keep some variable constant when we consider partial derivatives, as for example for $y(u,v)=\cos (2u)+v^2$ we write

$$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\frac{\partial u}{dx}+\frac{\partial y}{\partial v}\frac{\partial v}{dx}$$

with

  • $\frac{\partial y}{\partial u}=-2\sin (2u)$
  • $\frac{\partial y}{\partial v}=2v$
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I don't understand your solution. The derivative is $-sin(2u)\times 2\times u'$ by the chain rule. If you put $u=3x+1$ then you will get $-6sin(6x+2)$.

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