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Question Number 12155 by tawa last updated on 15/Apr/17

Find the 35th derivative of (2x^3 + 5x^4 )^(60)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{35th}\:\mathrm{derivative}\:\mathrm{of}\:\:\left(\mathrm{2x}^{\mathrm{3}} \:+\:\mathrm{5x}^{\mathrm{4}} \right)^{\mathrm{60}} \\ $$

Answered by mrW1 last updated on 15/Apr/17

y=(2x^3 +5x^4 )^(60) =Σ_(k=0) ^(60) C_k ^(60) (2x^3 )^k (5x^4 )^(60−k) y=Σ_(k=0) ^(60) C_k ^(60) ×2^k ×5^(60−k) ×x^(240−k) y^((1)) =Σ_(k=0) ^(60) C_k ^(60) ×2^k ×5^(60−k) ×(240−k)×x^(239−k) y^((2)) =Σ_(k=0) ^(60) C_k ^(60) ×2^k ×5^(60−k) ×(240−k)×(239−k)×x^(238−k) ...... y^((35)) =Σ_(k=0) ^(60) C_k ^(60) ×2^k ×5^(60−k) ×(240−k)×(239−k)×...×(206−k)×x^(205−k) y^((35)) =Σ_(k=0) ^(60) C_k ^(60) ×2^k ×5^(60−k) ×(((240−k)!)/((205−k)!))×x^(205−k) y^((35)) =Σ_(k=0) ^(60) A_k ×x^(205−k) A_k =C_k ^(60) ×2^k ×5^(60−k) ×(((240−k)!)/((205−k)!)) (k=0...60) A_0 =C_0 ^(60) ×2^0 ×5^(60−0) ×(((240−0)!)/((205−0)!))=5^(60) ×((240!)/(205!)) A_1 =C_1 ^(60) ×2^1 ×5^(60−1) ×(((240−1)!)/((205−1)!))=C_1 ^(60) ×2×5^(59) ×((239!)/(204!)) A_2 =C_2 ^(60) ×2^2 ×5^(60−2) ×(((240−2)!)/((205−2)!))=C_2 ^(60) ×2^2 ×5^(58) ×((238!)/(203!)) ......

$${y}=\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{4}} \right)^{\mathrm{60}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} \left(\mathrm{2}{x}^{\mathrm{3}} \right)^{{k}} \left(\mathrm{5}{x}^{\mathrm{4}} \right)^{\mathrm{60}−{k}} \\ $$$${y}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×{x}^{\mathrm{240}−{k}} \\ $$$${y}^{\left(\mathrm{1}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×\left(\mathrm{240}−{k}\right)×{x}^{\mathrm{239}−{k}} \\ $$$${y}^{\left(\mathrm{2}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×\left(\mathrm{240}−{k}\right)×\left(\mathrm{239}−{k}\right)×{x}^{\mathrm{238}−{k}} \\ $$$$...... \\ $$$${y}^{\left(\mathrm{35}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×\left(\mathrm{240}−{k}\right)×\left(\mathrm{239}−{k}\right)×...×\left(\mathrm{206}−{k}\right)×{x}^{\mathrm{205}−{k}} \\ $$$${y}^{\left(\mathrm{35}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×\frac{\left(\mathrm{240}−{k}\right)!}{\left(\mathrm{205}−{k}\right)!}×{x}^{\mathrm{205}−{k}} \\ $$$${y}^{\left(\mathrm{35}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{60}} {\sum}}{A}_{{k}} ×{x}^{\mathrm{205}−{k}} \\ $$$${A}_{{k}} =\mathrm{C}_{{k}} ^{\mathrm{60}} ×\mathrm{2}^{{k}} ×\mathrm{5}^{\mathrm{60}−{k}} ×\frac{\left(\mathrm{240}−{k}\right)!}{\left(\mathrm{205}−{k}\right)!}\:\:\:\:\:\:\left({k}=\mathrm{0}...\mathrm{60}\right) \\ $$$$ \\ $$$${A}_{\mathrm{0}} =\mathrm{C}_{\mathrm{0}} ^{\mathrm{60}} ×\mathrm{2}^{\mathrm{0}} ×\mathrm{5}^{\mathrm{60}−\mathrm{0}} ×\frac{\left(\mathrm{240}−\mathrm{0}\right)!}{\left(\mathrm{205}−\mathrm{0}\right)!}=\mathrm{5}^{\mathrm{60}} ×\frac{\mathrm{240}!}{\mathrm{205}!} \\ $$$${A}_{\mathrm{1}} =\mathrm{C}_{\mathrm{1}} ^{\mathrm{60}} ×\mathrm{2}^{\mathrm{1}} ×\mathrm{5}^{\mathrm{60}−\mathrm{1}} ×\frac{\left(\mathrm{240}−\mathrm{1}\right)!}{\left(\mathrm{205}−\mathrm{1}\right)!}=\mathrm{C}_{\mathrm{1}} ^{\mathrm{60}} ×\mathrm{2}×\mathrm{5}^{\mathrm{59}} ×\frac{\mathrm{239}!}{\mathrm{204}!} \\ $$$${A}_{\mathrm{2}} =\mathrm{C}_{\mathrm{2}} ^{\mathrm{60}} ×\mathrm{2}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{60}−\mathrm{2}} ×\frac{\left(\mathrm{240}−\mathrm{2}\right)!}{\left(\mathrm{205}−\mathrm{2}\right)!}=\mathrm{C}_{\mathrm{2}} ^{\mathrm{60}} ×\mathrm{2}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{58}} ×\frac{\mathrm{238}!}{\mathrm{203}!} \\ $$$$...... \\ $$

Commented by tawa last updated on 16/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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