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- Ceva’s Theorem Is More Than a Formula for Concurrency
- The Chain Rule Explained: Don't Just Memorize, Visualize It
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- Inverse Trigo
- Problems in progression
- General Solution of $y' = x \, \ln x$
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The differentiation you need…
The differentiation you need is direct from the following formulas:
From your equation, $u = 2x$ from which $du = 2$, hence,
$\dfrac{d}{dx} \left[ \tan 2x + \sec 2x \right] = 2 \sec^2 2x + 2 \sec 2x \tan 2x$