Question

22. If y = e^(2 log x + 3x)
, prove that dy/dx = x (2 + 3x) e^3x
​

Answer 1

Step-by-step explanation:

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)

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Answer 2

Answer:

dy/dx=x(2+3x)e^3x

Step-by-step explanation:

Given: y=e^(2 log x + 3x)

=> y = (e^2 log x)×(e^3x)

=> y = (x^2)×(e^3x) [since a^x=e^(x log a)]

Differentiating with respect to x, regarding y as a function of x, we get

dy/dx = d/dx [(x^2)×(e^3x)]

=> dy/dx = (d/dx (x^2))×(e^3x) + (d/dx(e^3x))×(x^2) [Product rule]

= (2x×(e^3x))+((x^2)×(e^3x)×d/dx(3x)) [Chain rule]

= (2x×(e^3x))+((x^2)×(e^3x)×3)

= x(2 + 3x) e^3x

Hence, proved.

This is the correct answer.