If f(x)=(x+1)(x4+x3+2)(x3+4x2+2x+3), then dydx∣∣∣x=−1 is
A
0
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B
2
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C
4
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D
8
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Solution
The correct option is D8 Let, derivative of x4+x3+2=L
and derivative of x3+4x2+2x+3=K ⇒f′(x)=(x+1)(L)(x3+4x2+2x+3)+(x+1)(x4+x3+2)(K)+(x4+x3+2)(x3+4x2+2x+3) ⇒f′(−1)=(2−1+1)(3−1−2+4)=(2)(4)∴f′(−1)=8