If y=(1+x)(1+2x)(1+3x), then the value of dydx at x=0 is
[1 mark]
A
5
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B
6
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C
2
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D
0
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Solution
The correct option is B6 y=(1+x)(1+2x)(1+3x)
At x=0,y=1
Taking ln on both sides, we get lny=ln(1+x)+ln(1+2x)+ln(1+3x)
Differentiating w.r.t. x, we get 1y⋅dydx=11+x+21+2x+31+3x ⇒dydx=y(11+x+21+2x+31+3x) ⇒dydx∣∣∣(0,1)=1(1+2+3)=6