differentiate with respect to x
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Given : y=(x−3)(x2+4)(◀+▶)−−−−−−−−−−−−−√.
Taking logarithm on both sides of (i), we get
logy=12{log(x−3)+log(x−3)+log(x2+4)−log(3x2+4x+5)}.
Differentiating both sides w.r.t. x, we get
1y.dydx=12.{1(x−3)+2x(x2+4)−(6x+4)(3x2+4x+5)}
⇒dydx=(12y).{1(x−3)+2x(x2+4)−(6x+4)(3x2+4x+5)}
=12.(x−3)(x2+4)(3x3+4x+4)−−−−−−−−−−−−−√.{1(x−3)+2x(x2+4)−(6x+4)(3x2+4x+5)}
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