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EXERCISE 1.5 · Q181

Q.e2x⋅tan⁡xe^{2x}\cdot\tan x

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✓ Free question

Let y=e2xtan⁡xy=e^{2x}\tan x.

Step 1 — first derivative by the product rule: dydx=2e2xtan⁡x+e2xsec⁡2x=e2x(2tan⁡x+sec⁡2x)\dfrac{dy}{dx}=2e^{2x}\tan x+e^{2x}\sec^2x=e^{2x}(2\tan x+\sec^2x).

Step 2 — differentiate again, treating e2xe^{2x} and (2tan⁡x+sec⁡2x)(2\tan x+\sec^2x) as the two factors. First note ddx(2tan⁡x+sec⁡2x)=2sec⁡2x+2sec⁡x⋅sec⁡xtan⁡x=2sec⁡2x+2sec⁡2xtan⁡x\dfrac{d}{dx}(2\tan x+\sec^2x)=2\sec^2x+2\sec x\cdot\sec x\tan x=2\sec^2x+2\sec^2x\tan x.

d2ydx2=2e2x(2tan⁡x+sec⁡2x)+e2x(2sec⁡2x+2sec⁡2xtan⁡x)\dfrac{d^2y}{dx^2}=2e^{2x}(2\tan x+\sec^2x)+e^{2x}(2\sec^2x+2\sec^2x\tan x)

Step 3 — expand and collect: =e2x[4tan⁡x+2sec⁡2x+2sec⁡2x+2sec⁡2xtan⁡x]=e2x[4tan⁡x+4sec⁡2x+2sec⁡2xtan⁡x]=e^{2x}[4\tan x+2\sec^2x+2\sec^2x+2\sec^2x\tan x]=e^{2x}[4\tan x+4\sec^2x+2\sec^2x\tan x].

d2ydx2=2e2x(2tan⁡x+2sec⁡2x+sec⁡2xtan⁡x)\dfrac{d^2y}{dx^2}=2e^{2x}\left(2\tan x+2\sec^2x+\sec^2x\tan x\right)

✓Final answer

d2ydx2=2e2x(2tan⁡x+2sec⁡2x+sec⁡2xtan⁡x)\dfrac{d^2y}{dx^2}=2e^{2x}\left(2\tan x+2\sec^2x+\sec^2x\tan x\right)

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