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  •  If   <img alt="y=\cot x+\sin x+\tan x+e^x \text { then find } \frac{d^3 y}{d x^3} ." src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D%5Ccot%20x&plus;%5Csin%20x&plus

 If   y=\cot x+\sin x+\tan x+e^x \text { then find } \frac{d^3 y}{d x^3} .

Option: 1

e^x-2 \operatorname{cosec}^3 x-4 \cos ^2 x \cdot \operatorname{cosec}^2 x-\cos x+4 \sec ^2 x \cdot \tan x+2 \sec ^2 x \mid


Option: 2

e^x-2 \operatorname{cosec}^3 x-4 \cot ^2 x \cdot \operatorname{cosec}^2 x-\cos x+4 \sec ^2 x \cdot \tan x+2 \sec ^4 x


Option: 3

e^x-2 \operatorname{cosec}^3 x-4 \cot ^2 x \cdot \operatorname{cosec}^2 x-\cot x+4 \sec ^2 x \cdot \tan x+2 \sec ^2 x


Option: 4

e^x-2 \operatorname{cosec}^3 x-4 \cot ^2 x \cdot \operatorname{cosec}^2 x-\cot x+4 \sec ^2 x \cdot \tan x+2 \sec ^4 x


Answers (1)

best_answer

Differentiate using trigonometric identities


\begin{aligned} & \frac{d y}{d x}=e^x-\operatorname{cosec}^2 x+\cos x+\sec ^2 x \\ & \frac{d^2 y}{d x^2}=e^x-\left[-2 \cot x \cdot \operatorname{cosec}^2 x\right]-\sin x+\left[2 \sec ^2 x \cdot \tan x\right] \\ & \frac{d^3 y}{d x^3}=e^x+2\left[-\operatorname{cosec}^2 x \cdot \operatorname{cosec}^2 x+-2 \cot x \cdot \operatorname{cosec}^2 x \cdot \cot x\right] \\ & -\cos x+2\left[2 \sec x \cdot \sec x \cdot \tan x+\sec ^2 x \cdot \sec ^2 x\right] \\ & \frac{d^3 y}{d x^3}=e^x+2\left[-\operatorname{cosec}^3 x-2 \cot ^2 x \cdot \operatorname{cosec}^2 x\right]-\cos x+2\left[2 \sec ^2 x \cdot \tan x+\sec ^4 x\right] \\ & \frac{d^3 y}{d x^3}=e^x-2 \operatorname{cosec}^3 x-4 \cot ^2 x \cdot \operatorname{cosec}^2 x-\cos x+4 \sec ^2 x \cdot \tan x+2 \sec ^4 x \\ & \end{aligned}

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