Derivatives of Hyperbolic Functions

The hyperbolic functions follow specific derivative patterns that you'll need to memorize:

$\frac{d}{dx}\sinh u = \cosh u \frac{du}{dx}$ $\frac{d}{dx}\cosh u = \sinh u \frac{du}{dx}$ $\frac{d}{dx}\tanh u = \text{sech}^2 u \frac{du}{dx}$ $\frac{d}{dx}\csc h u = -\csc h u \cot h u \frac{du}{dx}$ $\frac{d}{dx}\sec h u = -\sec h u \tan h u \frac{du}{dx}$ $\frac{d}{dx}\cot h u = -\csc h^2 u \frac{du}{dx}$

Unlike trigonometric functions where co-functions have negative derivatives, in hyperbolic functions, the reciprocal identities (csch, sech, coth) have the negative derivatives.

💡 Think of this as a pattern: When finding derivatives of hyperbolic functions, always look for the chain rule application by identifying the inner function u and its derivative du/dx.

Let's apply this to find the derivative of $y = \sinh 4x$:

1. Identify that $u = 4x$
2. Apply the formula: $\frac{dy}{dx} = \cosh 4x \frac{d}{dx}(4x) = \cosh 4x(4) = 4\cosh 4x$

Product Rule with Hyperbolic Functions

When dealing with products involving hyperbolic functions, you'll need to combine the product rule with the hyperbolic derivative formulas.

For the function $y = (1-x)^2 \sinh 2x$, we can identify:

• $u = (1-x)^2$
• $v = \sinh 2x$

Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$:

$\frac{dy}{dx} = (1-x)^2 \frac{d}{dx}(\sinh 2x) + \sinh 2x \frac{d}{dx}(1-x)^2$

For the first term, we apply the hyperbolic derivative: $\frac{d}{dx}(\sinh 2x) = \cosh 2x \cdot 2 = 2\cosh 2x$

For the second term, we need the power rule: $\frac{d}{dx}(1-x)^2 = 2(1-x)(-1) = -2(1-x)$

🔑 When working with complex hyperbolic expressions, break them down step by step, applying one rule at a time rather than trying to solve everything at once.

Combining everything: $\frac{dy}{dx} = 2(1-x)[(1-x)(\cosh 2x) - \sinh 2x]$

Combining Exponential and Hyperbolic Functions

Exponential and hyperbolic functions often appear together, as with $y = e^{-x}\cosh x$. Let's break this down:

• $u = e^{-x}$
• $v = \cosh x$

Using the product rule: $\frac{dy}{dx} = e^{-x}\frac{d}{dx}(\cosh x) + \cosh x \frac{d}{dx}(e^{-x})$

For the first part: $\frac{d}{dx}(\cosh x) = \sinh x$

For the second part: $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot (-1) = -e^{-x}$

This gives us: $\frac{dy}{dx} = e^{-x}\sinh x - e^{-x}\cosh x = e^{-x}(\sinh x - \cosh x)$

You can simplify further using hyperbolic function definitions: $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$

💡 Remember that hyperbolic functions are defined in terms of exponential functions, which makes them especially useful when both types appear in the same problem.

Through substitution and simplification, the final answer becomes: $\frac{dy}{dx} = -e^{-2x}$

Inverse Trigonometric Functions with Hyperbolic Arguments

When an inverse trigonometric function has a hyperbolic function as its argument, you'll need to combine multiple derivative rules. For $y = \arctan \sinh x$:

First, use the derivative formula for arctangent: $\frac{d}{dx}(\arctan u) = \frac{1}{1 + u^2}\frac{du}{dx}$

With $u = \sinh x$: $\frac{dy}{dx} = \frac{1}{1 + (\sinh x)^2}\frac{d}{dx}(\sinh x)$

Since $\frac{d}{dx}(\sinh x) = \cosh x$: $\frac{dy}{dx} = \frac{\cosh x}{1 + \sinh^2 x}$

This is where hyperbolic identities become powerful. Using $\cosh^2 x - \sinh^2 x = 1$, we can rearrange to $\cosh^2 x = 1 + \sinh^2 x$.

🔍 Hyperbolic identities can dramatically simplify expressions - knowing them well can turn complex derivatives into elegant solutions.

Substituting this identity: $\frac{dy}{dx} = \frac{\cosh x}{\cosh^2 x} = \frac{1}{\cosh x} = \text{sech } x$

Logarithmic Functions with Hyperbolic Arguments

When faced with logarithmic functions containing hyperbolic expressions, start by using logarithmic properties to simplify. For $y = \ln \tanh^2 3x$:

Using the property $\ln x^k = k \ln x$: $y = \ln \tanh^2 3x = 2 \ln \tanh 3x$

Now apply the logarithmic derivative formula: $\frac{d}{dx}(\ln u) = \frac{1}{u}\frac{du}{dx}$

$\frac{dy}{dx} = 2 \cdot \frac{1}{\tanh 3x} \cdot \frac{d}{dx}(\tanh 3x)$

Since $\frac{d}{dx}(\tanh u) = \text{sech}^2 u \frac{du}{dx}$: $\frac{dy}{dx} = 2 \cdot \frac{1}{\tanh 3x} \cdot \text{sech}^2 3x \cdot 3$

$= 6 \cdot \frac{\text{sech}^2 3x}{\tanh 3x}$

💡 When working with complex hyperbolic expressions, convert everything to basic hyperbolic functions (sinh and cosh) to simplify further.

Using the definition $\tanh x = \frac{\sinh x}{\cosh x}$ and $\text{sech}^2 x = \frac{1}{\cosh^2 x}$: $\frac{dy}{dx} = 6 \cdot \frac{1}{\cosh^2 3x} \cdot \frac{\cosh 3x}{\sinh 3x} = \frac{6}{\cosh 3x \sinh 3x}$

Hyperbolic Identities and Final Simplification

To further simplify expressions involving hyperbolic functions, you'll need to master hyperbolic identities. For our derivative $\frac{dy}{dx} = \frac{6}{\cosh 3x \sinh 3x}$:

We can use the identity $\sinh 2u = 2\sinh u \cosh u$ by multiplying numerator and denominator by 2: $\frac{dy}{dx} = \frac{6}{\cosh 3x \sinh 3x} \cdot \frac{2}{2} = \frac{12}{2\sinh 3x \cosh 3x} = \frac{12}{\sinh 6x}$

Using the reciprocal identity $\text{csch } x = \frac{1}{\sinh x}$: $\frac{dy}{dx} = 12 \text{ csch } 6x$

Key Hyperbolic Identities to Remember:

• Reciprocal identities: $\text{csch } x = \frac{1}{\sinh x}$, $\text{sech } x = \frac{1}{\cosh x}$, $\text{coth } x = \frac{1}{\tanh x}$
• Fundamental identity: $\cosh^2 x - \sinh^2 x = 1$
• Double angle formulas: $\sinh 2x = 2\sinh x \cosh x$, $\cosh 2x = \cosh^2 x + \sinh^2 x$
• Exponential forms: $\sinh x = \frac{e^x - e^{-x}}{2}$, $\cosh x = \frac{e^x + e^{-x}}{2}$

🌟 The beauty of hyperbolic functions is in their patterns. Once you understand their relationships, you can transform complicated expressions into elegant solutions!