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EXERCISE 1.5 · Q182

Q.e4x⋅cos⁡5xe^{4x}\cdot\cos 5x

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✓ Free question

Let y=e4xcos⁡5xy=e^{4x}\cos5x.

Step 1 — first derivative: dydx=4e4xcos⁡5x−5e4xsin⁡5x=e4x(4cos⁡5x−5sin⁡5x)\dfrac{dy}{dx}=4e^{4x}\cos5x-5e^{4x}\sin5x=e^{4x}(4\cos5x-5\sin5x).

Step 2 — differentiate again: d2ydx2=4e4x(4cos⁡5x−5sin⁡5x)+e4x(−20sin⁡5x−25cos⁡5x)\dfrac{d^2y}{dx^2}=4e^{4x}(4\cos5x-5\sin5x)+e^{4x}(-20\sin5x-25\cos5x).

Step 3 — expand and collect like terms: =e4x[16cos⁡5x−20sin⁡5x−20sin⁡5x−25cos⁡5x]=e4x[(16−25)cos⁡5x+(−40)sin⁡5x]=e^{4x}[16\cos5x-20\sin5x-20\sin5x-25\cos5x]=e^{4x}[(16-25)\cos5x+(-40)\sin5x].

d2ydx2=e4x(−9cos⁡5x−40sin⁡5x)=−e4x(9cos⁡5x+40sin⁡5x)\dfrac{d^2y}{dx^2}=e^{4x}(-9\cos5x-40\sin5x)=-e^{4x}(9\cos5x+40\sin5x)

✓Final answer

d2ydx2=−e4x(9cos⁡5x+40sin⁡5x)\dfrac{d^2y}{dx^2}=-e^{4x}(9\cos5x+40\sin5x)

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