Mathematical Problem & Solution

Mathematical Exploration

The Problem

We are asked to find the derivative of the function:

(ii) \quad y = \cot^{1}2x^3 + 1

The notation $\cot^{1}$ is often ambiguous. For this solution, we will interpret it as the inverse cotangent function (also written as $\operatorname{arccot}$ or $\cot^{-1}$). If a different interpretation was intended, the solution would change accordingly.

So, we interpret the function as:

y = \operatorname{arccot}(2x^3) + 1

The Solution: Finding $\frac{dy}{dx}$

To find the derivative $\frac{dy}{dx}$, we'll use the chain rule.

Step 1: Identify the outer and inner functions.

Let the inner function be $u = 2x^3$.

Then the outer function becomes $y = \operatorname{arccot}(u) + 1$.

Step 2: Differentiate the inner function.

The derivative of $u$ with respect to $x$ is:

\frac{du}{dx} = \frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2

Step 3: Differentiate the outer function.

The derivative of $y$ with respect to $u$ is:

Recall that $\frac{d}{du}(\operatorname{arccot}(u)) = -\frac{1}{1+u^2}$.

The derivative of the constant $1$ is $0$.

\frac{dy}{du} = \frac{d}{du}(\operatorname{arccot}(u) + 1) = -\frac{1}{1+u^2} + 0 = -\frac{1}{1+u^2}

Step 4: Apply the Chain Rule.

The chain rule states: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

\frac{dy}{dx} = \left(-\frac{1}{1+u^2}\right) \cdot (6x^2)

Step 5: Substitute back the inner function.

Substitute $u = 2x^3$ back into the expression:

\frac{dy}{dx} = -\frac{1}{1+(2x^3)^2} \cdot 6x^2

Simplify the expression:

\frac{dy}{dx} = -\frac{1}{1+4x^6} \cdot 6x^2

\frac{dy}{dx} = -\frac{6x^2}{1+4x^6}

Final Answer:

The derivative of $y = \operatorname{arccot}(2x^3) + 1$ is:

\boxed{\frac{dy}{dx} = -\frac{6x^2}{1+4x^6}}

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(ii) y=cot^ 12x^3+1