4 The curve $y=\mathrm{e}^{-4 x} an x$ has two stationary points in the interval $0 \leqslant x

Q: 4 The curve $y=\mathrm{e}^{-4 x} \tan x$ has two stationary points in the interval $0 \leqslant x<\frac{1}{2} \pi$. (a) Obtain an expression for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ and show it can be written in the form $\sec ^{2} x(a+b \sin 2 x) \mathrm{e}^{-4 x}$, where $a$ and $b$ are constants.

∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of $ y = e^{-4x} \tan(x) $, we use the product rule which states that $ (fg)' = f'g + fg' $. ‖ ‖ step 2 ⋮ The derivative of $ e^{-4x} $ is $ -4e^{-4x} $, and the derivative of $ \tan(x) $ is $ \sec^2(x) $. ‖ ‖ step 3 ⋮ Applying the product rule, we get $ \frac{dy}{dx} = e^{-4x} \sec^2(x) - 4e^{-4x} \tan(x) $. ‖ ‖ step 4 ⋮ We can express $ \tan(x) $ as $ \frac{\sin(2x)}{2} $ since $ \tan(x) = \frac{\sin(2x)}{2\cos(x)} $. ‖ ‖ step 5 ⋮ Substituting $ \tan(x) $ into the derivative, we get $ \frac{dy}{dx} = e^{-4x} \sec^2(x) - 2e^{-4x} \sin(2x) $. ‖ ‖ step 6 ⋮ We can factor out $ e^{-4x} \sec^2(x) $ to write the derivative in the form $ \sec^2(x)(a + b\sin(2x))e^{-4x} $. ‖ ‖ step 7 ⋮ Comparing the expression to the required form, we identify $ a = 1 $ and $ b = -2 $. ‖ ∻Answer∻ ⚹ $ \frac{dy}{dx} = \sec^2(x)(1 - 2\sin(2x))e^{-4x} $, where $ a = 1 $ and $ b = -2 $. ⚹ ∻Key Concept∻ ⚹Product Rule and Trigonometric Identities⚹ ∻Explanation∻ ⚹To find the derivative of a product of functions, we use the product rule. Additionally, trigonometric identities help us to express the derivative in a specific form, allowing us to identify the constants $ a $ and $ b $.⚹◊What is the significance of the two stationary points of the curve $y=e^{-4x}\tan x$ in the interval $0 \leqslant x

Question

4 The curve $y=\mathrm{e}^{-4 x} 	an x$ has two stationary points in the interval $0 \leqslant x<\frac{1}{2} \pi$.
(a) Obtain an expression for $\frac{\mathrm{d} y}{\mathrm{~d} x}$ and show it can be written in the form $\sec ^{2} x(a+b \sin 2 x) \mathrm{e}^{-4 x}$, where $a$ and $b$ are constants.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of $ y = e^{-4x} \tan(x) $, we use the product rule which states that $ (fg)' = f'g + fg' $. ‖ ‖ step 2 ⋮ The derivative of $ e^{-4x} $ is $ -4e^{-4x} $, and the derivative of $ \tan(x) $ is $ \sec^2(x) $. ‖ ‖ step 3 ⋮ Applying the product rule, we get $ \frac{dy}{dx} = e^{-4x} \sec^2(x) - 4e^{-4x} \tan(x) $. ‖ ‖ step 4 ⋮ We can express $ \tan(x) $ as $ \frac{\sin(2x)}{2} $ since $ \tan(x) = \frac{\sin(2x)}{2\cos(x)} $. ‖ ‖ step 5 ⋮ Substituting $ \tan(x) $ into the derivative, we get $ \frac{dy}{dx} = e^{-4x} \sec^2(x) - 2e^{-4x} \sin(2x) $. ‖ ‖ step 6 ⋮ We can factor out $ e^{-4x} \sec^2(x) $ to write the derivative in the form $ \sec^2(x)(a + b\sin(2x))e^{-4x} $. ‖ ‖ step 7 ⋮ Comparing the expression to the required form, we identify $ a = 1 $ and $ b = -2 $. ‖ ∻Answer∻ ⚹ $ \frac{dy}{dx} = \sec^2(x)(1 - 2\sin(2x))e^{-4x} $, where $ a = 1 $ and $ b = -2 $. ⚹ ∻Key Concept∻ ⚹Product Rule and Trigonometric Identities⚹ ∻Explanation∻ ⚹To find the derivative of a product of functions, we use the product rule. Additionally, trigonometric identities help us to express the derivative in a specific form, allowing us to identify the constants $ a $ and $ b $.⚹◊What is the significance of the two stationary points of the curve $y=e^{-4x}\tan x$ in the interval $0 \leqslant x<\frac{1}{2}\pi$ from a mathematical perspective?⍭ Generate me a similar question◊

Sia