Question

Evaluate the following derivatives: a. \(\frac{d}{d x}\left(\tanh \left(x^{2}+3 x\right)\right)\) b. \(\frac{d}{d x}\left(\frac{1}{(\sinh x)^{2}}\right)\)

Short answer

a. \( \sech^2(x^2 + 3x) \cdot (2x + 3) \); b. \( \frac{-2 \cosh x}{(\sinh x)^3} \)."}

Step-by-step solution

Identify the function for part (a)

The function to differentiate is given as \( \tanh(x^2 + 3x) \). We will need to apply the chain rule to differentiate this composite function.

Differentiate using the Chain Rule (part a)

The derivative of \( \tanh(u) \) with respect to \( u \) is \( \sech^2(u) \). Since \( u = x^2 + 3x \), we use the chain rule: \[ \frac{d}{dx}\left(\tanh(x^2 + 3x)\right) = \sech^2(x^2 + 3x) \cdot \frac{d}{dx}(x^2 + 3x) \]

Differentiate the inner function (part a)

Now, differentiate the inner function \( x^2 + 3x \): \[ \frac{d}{dx}(x^2 + 3x) = 2x + 3 \]

Combine results for part (a)

Plug the derivative of the inner function back into the expression from Step 2: \[ \frac{d}{dx}\left(\tanh(x^2 + 3x)\right) = \sech^2(x^2 + 3x) \cdot (2x + 3) \]

Identify the function for part (b)

The function is \( \frac{1}{(\sinh x)^2} \). Recognize this as \( (\sinh x)^{-2} \), which is a power function.

Apply the Chain Rule and Power Rule (part b)

Write \( \frac{1}{(\sinh x)^2} \) as \( (\sinh x)^{-2} \) and differentiate: \[ \frac{d}{dx}\left((\sinh x)^{-2}\right) = -2(\sinh x)^{-3} \cdot \frac{d}{dx}(\sinh x) \]

Differentiate \( \sinh x \) (part b)

The derivative of \( \sinh x \) is \( \cosh x \).

Combine results for part (b)

Replace \( \frac{d}{dx}(\sinh x) \) with \( \cosh x \) from Step 7: \[ \frac{d}{dx}\left((\sinh x)^{-2}\right) = -2(\sinh x)^{-3} \cdot \cosh x \] This results in: \( \frac{-2 \cosh x}{(\sinh x)^3} \).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that's crucial for finding derivatives of composite functions. In simpler terms, it's used when you have a function within another function. Imagine peeling an onion—you're tackling one layer at a time. Similarly, with the chain rule, you differentiate the outer function first and then multiply by the derivative of the inner function.

For example, let's consider the function \( f(x) = \tanh(x^2 + 3x) \). Here, \( x^2 + 3x \) is the inner function, and \( \tanh(u) \) where \( u = x^2 + 3x \) is the outer function.

• The derivative of the outer function \( \tanh(u) \) is \( \sech^2(u) \).
• Therefore, the derivative becomes \( \sech^2(x^2 + 3x) \cdot \frac{d}{dz}(x^2 + 3x) \).

This approach allows us to break down complex functions into manageable parts, making the calculus work much more straightforward.

Hyperbolic Functions

Hyperbolic functions, like hyperbolic sine \( \sinh \) and hyperbolic tangent \( \tanh \), are analogous to the trigonometric functions but for hyperbolas instead of circles. They have unique properties and derivatives useful in a variety of mathematical contexts, especially in calculus.

- The hyperbolic tangent function, \( \tanh(x) \), is defined as \( \frac{\sinh(x)}{\cosh(x)} \), and its derivative is \( \sech^2(x) \). This is similar to how the derivative of \( \tan(x) \) is \( \sec^2(x) \) in trigonometric functions.- The hyperbolic sine function, \( \sinh(x) \), has a derivative \( \cosh(x) \), paralleling the structure of sine and cosine in trigonometry.
These functions arise naturally in many areas such as engineering and physics, where they describe the shapes of hanging cables or the growth of certain phenomena. Understanding their derivatives is crucial for applying calculus to real-world problems that involve growth rates or rates of change in systems.

Power Rule

The power rule is one of the simplest yet most powerful tools in calculus for finding derivatives. It's particularly handy when dealing with polynomial and power functions. This rule states that to differentiate \( x^n \), you multiply by the exponent and reduce the power by one, producing \( nx^{n-1} \).

Now, when you meet compound expressions like \( \frac{1}{(\sinh x)^2} \), first rewrite it for simplicity as \( (\sinh x)^{-2} \). This makes it straightforward to apply the power rule.

• The expression becomes \( -2(\sinh x)^{-3} \), and then you multiply by the derivative of \( \sinh x \).
• Thus, the complete derivative is \( \frac{-2\cosh x}{(\sinh x)^3} \).

The power rule, as you can see, simplifies challenging expressions to make them more manageable, providing a quick and efficient route to finding derivatives.