2.6 Q-3

Question Statement

Find $\frac{d y}{d x}$ for the following functions:

1. $y = \cosh(2x)$
2. $y = \sinh(3x)$
3. $y = \tan^{-1}(\sin x)$
4. $y = \sinh^{-1}(x^3)$
5. $y = \ln(\tanh x)$
6. $y = \sinh^{-1}\left(\frac{x}{2}\right)$

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Background and Explanation

To solve these problems, we need to apply basic differentiation rules such as:

• Hyperbolic functions: Understand how to differentiate hyperbolic functions like $\cosh x$, $\sinh x$, and $\tanh x$.
• Inverse functions: Recognize that the derivatives of inverse hyperbolic functions like $\sinh^{-1}(x)$ and $\tanh^{-1}(x)$ involve a bit more manipulation.
• Basic differentiation: Apply standard derivative formulas for sine, cosine, and other elementary functions.

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Solution

1. $y = \cosh(2x)$

The derivative of $\cosh(2x)$ is:

$$\frac{d}{dx}(\cosh(2x)) = 2\sinh(2x)$$

Explanation:

• The chain rule is applied to differentiate the composition $\cosh(2x)$.
• The derivative of $\cosh(u)$ is $\sinh(u)$, and then multiply by the derivative of $2x$, which is $2$.

Answer:

$$\frac{d y}{d x} = 2 \sinh(2x)$$

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2. $y = \sinh(3x)$

The derivative of $\sinh(3x)$ is:

$$\frac{d}{dx}(\sinh(3x)) = 3\cosh(3x)$$

Explanation:

• The chain rule is applied again.
• The derivative of $\sinh(u)$ is $\cosh(u)$, and the derivative of $3x$ is $3$.

Answer:

$$\frac{d y}{d x} = 3 \cosh(3x)$$

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3. $y = \tan^{-1}(\sin x)$

We start by differentiating implicitly:

$$\tanhy = \sin x$$

Differentiating both sides with respect to $x$:

$$\sec^2 y \frac{dy}{dx} = \cos x$$

Now, solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{\cos x}{\sec^2 y} = \frac{\cos x}{1 - \tan^2 y}$$

Using the identity $\tan y = \sin x$, we can further simplify:

$$\frac{dy}{dx} = \frac{\cos x}{1 - \sin^2 x} = \frac{\cos x}{\cos^2 x} = \sec x$$

Answer:

$$\frac{d y}{d x} = \sec x$$

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4. $y = \sinh^{-1}(x^3)$

Start by differentiating the equation:

$$\sinh y = x^3$$

Differentiate both sides with respect to $x$:

$$\cosh y \frac{dy}{dx} = 3x^2$$

Now, solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{3x^2}{\cosh y}$$

Using the identity $\cosh^2 y = 1 + \sinh^2 y$ and substituting $\sinh y = x^3$, we get:

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

Answer:

$$\frac{d y}{d x} = \frac{3x^2}{\sqrt{1 + x^6}}$$

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5. $y = \ln(\tanh x)$

Differentiate using the chain rule:

$$\frac{dy}{dx} = \frac{1}{\tanh x} \cdot \frac{d}{dx}(\tanh x)$$

The derivative of $\tanh x$ is $\sech^2 x$, so:

$$\frac{dy}{dx} = \frac{\sech^2 x}{\tanh x}$$

Using the identity $\sech^2 x = 1 - \tanh^2 x$, the expression simplifies to:

$$\frac{dy}{dx} = \frac{2}{\sinh(2x)}$$

Answer:

$$\frac{d y}{d x} = 2 \csc h(2x)$$

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6. $y = \sinh^{-1}\left(\frac{x}{2}\right)$

Start with the equation:

$$\sinh y = \frac{x}{2}$$

Differentiate both sides with respect to $x$:

$$\cosh y \frac{dy}{dx} = \frac{1}{2}$$

Now, solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{1}{2 \cosh y}$$

Using the identity $\cosh^2 y = 1 + \sinh^2 y$ and $\sinh y = \frac{x}{2}$, we get:

$$\frac{dy}{dx} = \frac{1}{2 \sqrt{1 + \left(\frac{x}{2}\right)^2}} = \frac{1}{\sqrt{4 + x^2}}$$

Answer:

$$\frac{d y}{d x} = \frac{1}{\sqrt{4 + x^2}}$$

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Key Formulas or Methods Used

• Chain Rule: Used to differentiate composite functions.
• Hyperbolic Function Derivatives:
  • $\frac{d}{dx}(\cosh x) = \sinh x$
  • $\frac{d}{dx}(\sinh x) = \cosh x$
  • $\frac{d}{dx}(\tanh x) = \sech^2 x$
• Inverse Function Derivatives:
  • $\frac{d}{dx}(\sinh^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}}$
  • $\frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2}$

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Summary of Steps

1. Apply the chain rule where applicable (e.g., for $\cosh(2x)$, $\sinh(3x)$).
2. Use inverse hyperbolic function derivatives for functions like $\tan^{-1}(\sin x)$ and $\sinh^{-1}(x^3)$.
3. Simplify the expression using standard identities for hyperbolic functions.
4. Solve for $\frac{dy}{dx}$ by rearranging terms and applying algebraic simplifications.