Y=(2logx+3x)to the power of e. Then find the dy/dx.
Answers (2)
Y=e^(2logx+3x)
=e^(logx^2).e^3x
=x^2.e^3x
dy/dx = 2x.e^3x+x^2.3.e^3x
=2x.e^3x+3x^2.e^3x
:)
=e^(logx^2).e^3x
=x^2.e^3x
dy/dx = 2x.e^3x+x^2.3.e^3x
=2x.e^3x+3x^2.e^3x
:)
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Student Expert
21st Apr, 2019
Hey Sidh,
You want to know the differentiation of
Y = e^(2logx + 3x)
Taking log both sides
Log (y) = loge^ (2logx + 3x)
Log (y) = (2logx + 3x) log e
Log (y) = 2logx + 3x
Differentiating both sides with respect to x
(1 / y) (dy/dx) = 2 / x + 3
dy/dx = (2/x + 3) y
dy/dx = (2/x + 3)(2logx +3x)
Hope it helps :)
You want to know the differentiation of
Y = e^(2logx + 3x)
Taking log both sides
Log (y) = loge^ (2logx + 3x)
Log (y) = (2logx + 3x) log e
Log (y) = 2logx + 3x
Differentiating both sides with respect to x
(1 / y) (dy/dx) = 2 / x + 3
dy/dx = (2/x + 3) y
dy/dx = (2/x + 3)(2logx +3x)
Hope it helps :)
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