Differentiate trigonometric function Member for 3 years 10 months Submitted by perfectveneer777 on Tue, 06/07/2022 - 18:45 Hello everyone! I need some help again. Can anyone derive this problem with steps? d/dx ((tan2x)+(sec2x)) Thank you. Tags Derivation Trigonometric Function Log in or register to post comments The differentiation you need… Member for 17 years 6 months Jhun Vert Wed, 07/24/2024 - 10:29 The differentiation you need is direct from the following formulas: $d(\tan u) = \sec^2 u ~ du$ $d(\sec u) = \sec u \tan u ~ du$ From your equation, $u = 2x$ from which $du = 2$, hence, $\dfrac{d}{dx} \left[ \tan 2x + \sec 2x \right] = 2 \sec^2 2x + 2 \sec 2x \tan 2x$ Log in or register to post comments
The differentiation you need… Member for 17 years 6 months Jhun Vert Wed, 07/24/2024 - 10:29 The differentiation you need is direct from the following formulas: $d(\tan u) = \sec^2 u ~ du$ $d(\sec u) = \sec u \tan u ~ du$ From your equation, $u = 2x$ from which $du = 2$, hence, $\dfrac{d}{dx} \left[ \tan 2x + \sec 2x \right] = 2 \sec^2 2x + 2 \sec 2x \tan 2x$ Log in or register to post comments
The differentiation you need…
Member for
17 years 6 monthsThe differentiation you need is direct from the following formulas:
From your equation, $u = 2x$ from which $du = 2$, hence,
$\dfrac{d}{dx} \left[ \tan 2x + \sec 2x \right] = 2 \sec^2 2x + 2 \sec 2x \tan 2x$