Question

Derivatives of tower functions (or \(g^{h}\) ) Find the derivative of each function and evaluate the derivative at the given value of \(a\). $$f(x)=(\tan x)^{x-1} ; a=\pi / 4$$

Short answer

Based on the given function \(f(x)=(\tan x)^{x-1}\), find the derivative and evaluate it at \(a = \frac{\pi}{4}\). Solution: The derivative of the function evaluated at \(a = \frac{\pi}{4}\) is \(\frac{\pi - 2}{2}\).

Step-by-step solution

Find the derivative of g(x) and h(x)

We need to find the first derivatives of g(x) and h(x). In this case, \(g(x) = \tan x\) and \(h(x) = x - 1\). So, $$g'(x) = \frac{d}{dx} \tan (x) = \sec^2(x)$$ $$h'(x) = \frac{d}{dx} (x - 1) = 1$$

Apply the chain rule

Now we'll apply the chain rule, which says that the derivative of a composite function \([(g \circ h)(x) = g(h(x))]\) is the product of the derivative of the outer function g with respect to h, and h with respect to x. In our case \(g(x)=\tan x\) and \(h(x) = x - 1\), the chain rule for finding the derivative of the function \((\tan x)^{x-1}\) is: $$\frac{d}{dx}(\tan x)^{x-1} = (x-1) \frac{d}{dx}[\tan^{x-1}(x)]$$

Apply the product rule

Use the product rule to find the derivative of the function \((x-1)(\tan^{x-1}(x))\). The product rule states that if we have a function which is a product of two functions, say \(u(x)\) and \(v(x)\), the derivative is: $$\frac{d}{dx}[u(x)v(x)] = u'(x) v(x) + u(x) v'(x)$$ Applying the product rule, we get $$ \frac{d}{dx}(\tan x)^{x-1} = (\tan^{x-1}(x))[(x-1)\sec^2(x) + \tan^{x-1}(x)]$$ Now, this will be our derivative function for \(f(x)\).

Evaluate the derivative at a = π/4

Substitute \(a = \frac{\pi}{4}\) in the derivative function we got in step 3. $$f'(\frac{\pi}{4}) = (\tan^{\frac{\pi}{4}-1}(\frac{\pi}{4}))[(\frac{\pi}{4}-1)\sec^2(\frac{\pi}{4}) + \tan^{\frac{\pi}{4}-1}(\frac{\pi}{4})]$$ Since, \(\tan(\frac{\pi}{4}) = 1\) and \(\sec(\frac{\pi}{4}) = \sqrt{2}\), plug in these values to get the final answer: $$f'(\frac{\pi}{4}) = (1)[(\frac{\pi}{4}-1)2 + 1]$$ $$f'(\frac{\pi}{4}) = \frac{\pi - 2}{2}$$ So the derivative of the given function evaluated at \(a = \frac{\pi}{4}\) is \(\frac{\pi - 2}{2}\).

Key concepts

Chain Rule

The chain rule is fundamental when differentiating compositions of functions. In simple terms, it helps us find the derivative of a function that is the result of one function applied to another. Consider two functions, where one is inside the other, such as \(g(h(x))\). To find the derivative of this composition, use the chain rule:
\(\frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x)\).
The key is to differentiate the outer function first, then multiply by the derivative of the inner function. This process allows you to "chain" the different parts of the composition together, handling each function in its natural order.In our exercise, \(f(x) = (\tan x)^{x-1}\) needs the chain rule. Here, \(g(x) = (\tan x)\), and \(h(x) = x - 1\). By the chain rule's principle, we derive the outer part \(x-1\) while keeping \(\tan x\) unchanged, then differentiate \(\tan x\) to find the overall derivative.

Product Rule

The product rule comes into play when differentiating products of two functions. If you have a function in the form of \(u(x) \cdot v(x)\), the product rule states:
\(\frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x)\).
This rule allows you to handle the derivative of each function separately and then combine them by multiplying and adding.In our problem, we arrive at needing the product rule because we express the derivative as a product of \(x-1\) and \(\tan^{x-1}(x)\). Differentiating each piece gives us the derivative of the whole. This approach aligns each function's derivative in a natural way, making it easier to integrate them into the existing expression.

Trigonometric Derivatives

Trigonometric derivatives are important for understanding how trigonometric functions change. These derivatives allow you to express rates of change for trigonometric functions like sine, cosine, and tangent.For tangent, the derivative is:
\(\frac{d}{dx} \tan(x) = \sec^2(x)\).
This derivative tells us that the rate of change of \(\tan(x)\) is proportional to \(\sec(x)\) squared, an essential piece of information when dealing with functions like \(\tan(x)\), as in our exercise.By understanding how these derivatives work, you gain the power to tackle more sophisticated problems involving trigonometric functions and their compositions. This knowledge is key for solving our given problem, where \(\tan(x)\) plays a central role in the differentiation process.