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Chapter 6: Problem 12

If \(y=\left(1+\tan \left(\frac{\pi}{8}-x\right)\right)\left(1+\tan \left(x+\frac{\pi}{8}\right)\right)\), find \(\frac{d y}{d x}\)

Short Answer

Expert verified
\[ \frac{dy}{dx} = sec^2(x+\frac{\pi}{8})(1+\tan (\frac{\pi}{8} - x)) - (1+\tan \left(\frac{\pi}{8}-x\right)sec^2(\frac{\pi}{8}-x) \]

Step by step solution

01

Apply the Product Rule

Differentiate the expression by applying the product rule \( (uv)' = u'v + uv' \). Here, \( u = 1+\tan \left(\frac{\pi}{8}-x\right) \) and \( v = 1+\tan \left(x+\frac{\pi}{8}\right) \). We therefore get: \(du/dx = sec^2(\frac{\pi}{8}-x) * -1\) (use of chain rule) and \(dv/dx = sec^2(x+\frac{\pi}{8})\). Therefore, \( (dy/dx) = u'v + uv' \).
02

Substitute and Simplify

Substitute \(u, du/dx, v \) and \(dv/dx \) in the equation \( (dy/dx) = u'v + uv' \). After simplification, we have \( \frac{dy}{dx} = -(1+\tan \left(\frac{\pi}{8}-x\right))sec^2(\frac{\pi}{8}-x) + sec^2(x+\frac{\pi}{8})(1+\tan (\frac{\pi}{8} - x)) \). Remember to distribute correctly.
03

Rearrange Terms

Rearrange the equation to obtain a simpler form. The answer becomes more clear in the form \[ \frac{dy}{dx} = sec^2(x+\frac{\pi}{8})(1+\tan (\frac{\pi}{8} - x)) - (1+\tan \left(\frac{\pi}{8}-x\right)sec^2(\frac{\pi}{8}-x) \].

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