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I want derivative for (x+1)^2×e^x
please help​

User Chindraba
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1 Answer

3 votes

Answer:

(x + 1)(x + 3)eˣ

Explanation:

Since f(x) = (x + 1)²eˣ

df(x)/dx = d[(x + 1)²eˣ]/dx

By the product rule, duv/dx = udv/dx + vdu/dx

where u = (x + 1)² and v = eˣ

So, df(x)/dx = d[(x + 1)²eˣ]/dx

= (x + 1)²deˣ/dx + eˣd(x + 1)²/dx

= (x + 1)²eˣ + eˣd(x + 1)²/d(x + 1) × d(x + 1)/dx (by the chain-rule dy/dx = dy/du × du/dx)

= (x + 1)²eˣ + eˣ[2(x + 1) × 1]

= (x + 1)²eˣ + 2eˣ(x + 1)

= (x + 1)eˣ(x + 1 + 2)

= (x + 1)eˣ(x + 3)

= (x + 1)(x + 3)eˣ

User Shamdor
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