answer this in terms of invers trig functions differentiation: y= arctan(1/t)

To differentiate the function \( y = \arctan\left( \frac{1}{t} \right) \) with respect to \( t \), we can apply the chain rule along with the known derivative formula for \( \arctan(x) \).

### Derivative of \( \arctan(x) \):

The derivative of \( \arctan(x) \) with respect to \( x \) is:

\[
\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}
\]

Now, apply this to the function \( y = \arctan\left( \frac{1}{t} \right) \).

1. First, recognize that \( x = \frac{1}{t} \).
2. We apply the chain rule, meaning we differentiate \( \arctan(x) \) with respect to \( x \), and then multiply by the derivative of \( x = \frac{1}{t} \) with respect to \( t \).

### Step 1: Differentiate \( \arctan(x) \) with respect to \( x \):

\[
\frac{d}{dx} \arctan\left( \frac{1}{t} \right) = \frac{1}{1 + \left( \frac{1}{t} \right)^2} = \frac{1}{1 + \frac{1}{t^2}} = \frac{t^2}{t^2 + 1}
\]

### Step 2: Differentiate \( \frac{1}{t} \) with respect to \( t \):

The derivative of \( \frac{1}{t} \) is:

\[
\frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}
\]

### Step 3: Apply the chain rule:

Now, multiply the two derivatives:

\[
\frac{dy}{dt} = \frac{d}{dx} \arctan\left( \frac{1}{t} \right) \cdot \frac{d}{dt} \left( \frac{1}{t} \right)
\]

\[
\frac{dy}{dt} = \left( \frac{t^2}{t^2 + 1} \right) \cdot \left( -\frac{1}{t^2} \right)
\]

Simplifying:

\[
\frac{dy}{dt} = -\frac{1}{t^2 + 1}
\]

### Final Answer:
\[
\frac{dy}{dt} = -\frac{1}{t^2 + 1}
\]

Would you like further details on the differentiation process or have any questions? Here's some related material you might explore:

1. How to differentiate other inverse trigonometric functions like \( \arcsin(x) \) and \( \arccos(x) \).
2. The chain rule in differentiation: when and how to apply it.
3. How to apply the quotient rule in differentiation.
4. The geometry of the inverse tangent function and its derivatives.
5. The relationship between trigonometric functions and their inverses.

**Tip:** Always check the domain of the inverse trigonometric function you're differentiating, as certain values might make the derivative undefined.

Learn how to differentiate the function y = arctan(1/t) with respect to t using the chain rule. This step-by-step solution demonstrates key concepts of differentiation, inverse trigonometric functions, and applying the chain rule. Suitable for students in Grades 11-12.

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