Derivatives of Hyperbolic Functions — Question 7

Question 7 - Solution

We are given: $$f(x) = \tanh(x^2) \cdot \text{sech}(x)$$

(a) Use the product rule:

Let: $$u = \tanh(x^2), \quad v = \text{sech}(x)$$

Then: $$f'(x) = u'v + uv'$$

Differentiate each part:

Step 1: $u = \tanh(x^2) \Rightarrow u' = \text{sech}^2(x^2) \cdot 2x$ using chain rule

Step 2: $v = \text{sech}(x) \Rightarrow v' = -\text{sech}(x) \cdot \tanh(x)$

Now substitute into the product rule:

$$f'(x) = \left[ \text{sech}^2(x^2) \cdot 2x \right] \cdot \text{sech}(x) + \tanh(x^2) \cdot \left[ -\text{sech}(x) \cdot \tanh(x) \right]$$

(b) Final expression: $$\boxed{ f'(x) = 2x\, \text{sech}^2(x^2)\, \text{sech}(x) - \tanh(x^2)\, \text{sech}(x)\, \tanh(x) }$$