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Question

If y=[x2+1x+1], then find dydx.

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Solution

Here, u(x)=x2+1,v(x)=x+1.
Using quotient rule, we get
dydx=(x+1)d(x2+1)dx(x2+1)d(x+1)dx(x+1)2
=(x+1)2x(x2+1)1(x+1)2
=2x2+2xx21(x+1)2=x2+2x1(x+1)2.

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