NTA Abhyas JEE Main2020MathematicsDifferentiationPractice

Let y = x l o g e x . If the value of d y d x at x = e 4 is k , then the value of 4 e 3 k is (use e = 2.7 )

Correct answer

13.5

Step-by-step solution

d y d x = 1 2 x l o g e x d d x x l o g e x = 1 2 x l o g e x x × 1 x + 1 × l o g e x ⇒ d y d x x = e 4 = 1 2 e 4 × 4 1 + 4 = 5 4 e 2 ∵ l o g e e = 1 Hence, 4 e 3 k = 4 e 3 5 4 e 2 = 5 e = 13 . 5

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