2.5 Q-5

Question Statement

Differentiate the following expressions with respect to the specified variables:

i. Differentiate $\sin x$ with respect to $\cot x$.

ii. Differentiate $\sin^2 x$ with respect to $\cos^4 x$.

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Background and Explanation

In this problem, we are asked to find the derivatives of trigonometric functions with respect to other trigonometric functions. To solve these problems, we will use:

• Basic Derivatives: The derivative of $\sin x$ is $\cos x$ and the derivative of $\cot x$ is $-\csc^2 x$.
• Chain Rule: When differentiating a composite function, we apply the chain rule, which states: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

The chain rule allows us to differentiate functions like $\sin x$ with respect to $\cot x$, or $\sin^2 x$ with respect to $\cos^4 x$.

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Solution

i. Differentiate $\sin x$ with respect to $\cot x$

We are given the function $\sin x$ and asked to differentiate it with respect to $\cot x$. We will apply the chain rule to solve this.

1. Define the functions:

   • Let $y = \sin x$.
   • Let $u = \cot x$.

2. Apply the chain rule: $\frac{d y}{d x} = \cos x$ (The derivative of $\sin x$ is $\cos x$). $\frac{d u}{d x} = -\csc^2 x$ (The derivative of $\cot x$ is $-\csc^2 x$).

3. Now, to differentiate $y$ with respect to $u$, we use the chain rule: $\frac{d y}{d u} = \frac{d y}{d x} \cdot \frac{d x}{d u}$

   Using the formula from the chain rule: $\frac{d x}{d u} = \frac{1}{\csc^2 x} = \sin^2 x$

4. Finally, substitute back into the chain rule: $\frac{d y}{d u} = \cos x \cdot (-\sin^2 x)$

   This simplifies to: $\frac{d y}{d u} = -\cos x \sin^2 x$

   Answer: $\boxed{-\cos x \sin^2 x}$

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ii. Differentiate $\sin^2 x$ with respect to $\cos^4 x$

We are now asked to differentiate $\sin^2 x$ with respect to $\cos^4 x$. Again, we will use the chain rule.

1. Define the functions:

   • Let $y = \sin^2 x$.
   • Let $u = \cos^4 x$.

2. Differentiate $\sin^2 x$: $\frac{d y}{d x} = 2 \sin x \cdot \cos x$ (Using the chain rule for $\sin^2 x$).

3. Differentiate $\cos^4 x$: $\frac{d u}{d x} = 4 \cos^3 x \cdot (-\sin x)$ (Using the chain rule for $\cos^4 x$).

4. Now, apply the chain rule: $\frac{d y}{d u} = \frac{d y}{d x} \cdot \frac{d x}{d u}$

   Substituting the expressions for $\frac{d y}{d x}$ and $\frac{d u}{d x}$: $\frac{d y}{d u} = 2 \sin x \cdot \cos x \cdot \frac{1}{4 \cos^3 x \cdot \sin x}$

5. Simplify the expression: $\frac{d y}{d u} = \frac{-1}{2 \cos^2 x}$

   Answer: $\boxed{-\frac{1}{2 \cos^2 x}}$

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Key Formulas or Methods Used

• Basic Derivatives:
  • $\frac{d}{dx} (\sin x) = \cos x$
  • $\frac{d}{dx} (\cot x) = -\csc^2 x$
  • $\frac{d}{dx} (\cos^4 x) = 4 \cos^3 x (-\sin x)$
• Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

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Summary of Steps

1. i. Differentiate $\sin x$ with respect to $\cot x$:

   • Find $\frac{d y}{d x}$ and $\frac{d u}{d x}$.
   • Apply the chain rule to get $\frac{d y}{d u} = -\cos x \sin^2 x$.

2. ii. Differentiate $\sin^2 x$ with respect to $\cos^4 x$:

   • Find $\frac{d y}{d x}$ and $\frac{d u}{d x}$.
   • Apply the chain rule to get $\frac{d y}{d u} = -\frac{1}{2 \cos^2 x}$.

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