Given that y = Sec'x. a) Show that d²y/dx² = 6Sec⁴x − 4Sec²x. b) Find a Taylor series expansion of Sec²x in ascending powers of (x - π/4), up to and including the term in (x - π/4)³.

Given $y = \sec^2 x = (\sec x)^2$.
Using the chain rule, $\frac{dy}{dx} = 2(\sec x) \cdot \frac{d}{dx}(\sec x)$.
Since $\frac{d}{dx}(\sec x) = \sec x \tan x$:

$$\frac{dy}{dx} = 2\sec x (\sec x \tan x) = 2\sec^2 x \tan x$$

Let $u = 2\sec^2 x$ and $v = \tan x$.
Then $\frac{du}{dx} = 2 \cdot (2\sec x \cdot \sec x \tan x) = 4\sec^2 x \tan x$.
And $\frac{dv}{dx} = \sec^2 x$.
Applying the product rule, $\frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx}$:

$$\frac{d^2y}{dx^2} = (2\sec^2 x)(\sec^2 x) + (\tan x)(4\sec^2 x \tan x)$$ $$\frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x \tan^2 x$$

Using the trigonometric identity $\tan^2 x = \sec^2 x - 1$:

$$\frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x (\sec^2 x - 1)$$ $$\frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^4 x - 4\sec^2 x$$ $$\frac{d^2y}{dx^2} = 6\sec^4 x - 4\sec^2 x$$

This matches the required expression.

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$

Here, $f(x) = \sec^2 x$ and $a = \frac{\pi}{4}$.

We know $\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\tan(\frac{\pi}{4}) = 1$.

$$f(x) = \sec^2 x \implies f\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = (\sqrt{2})^2 = 2$$

From part (a), we have the derivatives:

$$f'(x) = 2\sec^2 x \tan x \implies f'\left(\frac{\pi}{4}\right) = 2\sec^2\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) = 2(2)(1) = 4$$ $$f''(x) = 6\sec^4 x - 4\sec^2 x \implies f''\left(\frac{\pi}{4}\right) = 6\sec^4\left(\frac{\pi}{4}\right) - 4\sec^2\left(\frac{\pi}{4}\right) = 6(\sqrt{2})^4 - 4(\sqrt{2})^2 = 6(4) - 4(2) = 24 - 8 = 16$$

Now, find the third derivative:

$$f'''(x) = \frac{d}{dx}(6\sec^4 x - 4\sec^2 x)$$ $$f'''(x) = 6(4\sec^3 x \cdot \sec x \tan x) - 4(2\sec x \cdot \sec x \tan x)$$ $$f'''(x) = 24\sec^4 x \tan x - 8\sec^2 x \tan x$$ $$f'''\left(\frac{\pi}{4}\right) = 24\sec^4\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) - 8\sec^2\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right)$$ $$f'''\left(\frac{\pi}{4}\right) = 24(\sqrt{2})^4 (1) - 8(\sqrt{2})^2 (1) = 24(4) - 8(2) = 96 - 16 = 80$$

$$f(x) = 2 + 4\left(x-\frac{\pi}{4}\right) + \frac{16}{2}\left(x-\frac{\pi}{4}\right)^2 + \frac{80}{6}\left(x-\frac{\pi}{4}\right)^3 + \dots$$ $$f(x) = 2 + 4\left(x-\frac{\pi}{4}\right) + 8\left(x-\frac{\pi}{4}\right)^2 + \frac{40}{3}\left(x-\frac{\pi}{4}\right)^3 + \dots$$

The Taylor series expansion up to and including the term in $(x - \frac{\pi}{4})^3$ is:

$$\boxed{2 + 4\left(x-\frac{\pi}{4}\right) + 8\left(x-\frac{\pi}{4}\right)^2 + \frac{40}{3}\left(x-\frac{\pi}{4}\right)^3}$$