Find dy/dx at x=0 given y=1/(3+u^2) and u=2x+1.


We are given two equations:
- \( y = \frac{1}{3 + u^2} \)
- \( u = 2x + 1 \)

To find \(\frac{dy}{dx}\) at \(x = 0\), we apply the chain rule since \(y\) is a function of \(u\) and \(u\) is a function of \(x\).

### Step 1: Differentiate \(y\) with respect to \(u\)
Using the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
\]

First, differentiate \(y = \frac{1}{3 + u^2}\) with respect to \(u\):
\[
\frac{dy}{du} = \frac{d}{du} \left( \frac{1}{3 + u^2} \right)
\]
Using the quotient rule or power rule, we get:
\[
\frac{dy}{du} = -\frac{2u}{(3 + u^2)^2}
\]

### Step 2: Differentiate \(u\) with respect to \(x\)
Now, differentiate \(u = 2x + 1\) with respect to \(x\):
\[
\frac{du}{dx} = 2
\]

### Step 3: Apply the chain rule
Now apply the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{2u}{(3 + u^2)^2} \cdot 2
\]
\[
\frac{dy}{dx} = -\frac{4u}{(3 + u^2)^2}
\]

### Step 4: Evaluate at \(x = 0\)
At \(x = 0\), we find \(u\):
\[
u = 2(0) + 1 = 1
\]

Substitute \(u = 1\) into \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = -\frac{4(1)}{(3 + 1^2)^2} = -\frac{4}{(3 + 1)^2} = -\frac{4}{16} = -\frac{1}{4}
\]

Thus, \(\frac{dy}{dx}\) at \(x = 0\) is \(-\frac{1}{4}\).

Would you like more details on the steps or further clarifications?

Here are 5 related questions:
1. How does the chain rule work when you have multiple variables?
2. Can you explain the quotient rule and when it applies?
3. How would the solution change if \(u = 3x + 2\)?
4. How do we interpret \(\frac{dy}{dx}\) in terms of rates of change?
5. How would we approach finding higher-order derivatives of \(y\)?

**Tip:** Always substitute values after differentiation, not before, to avoid calculation errors.

Find dy/dx at x=0 given y=1/(3+u^2) and u=2x+1.

Learn how to find dy/dx at x=0 using the chain rule, given y = 1/(3 + u^2) and u = 2x + 1. This detailed step-by-step solution applies key calculus concepts such as the chain rule and quotient rule. Suitable for advanced high school students in Grades 11-12.

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