If x^y=e^x-y, then dy/ dx at x=1 is

Question

Accepted Answer

Correct answer: C. (c) : We have, x^y=e^x-y Taking logarithm on both sides, we get y log ⁡x=x-y ⇒y ( log ⁡x+1 )=x ⇒y= x log ⁡x+1 ⇒ dy dx= 1 (1+ log ⁡x )- 1x x (1+ log ⁡x )^2 ⇒ dy dx= 1+ log ⁡x-1 (1+ log ⁡x )^2= log ⁡x (1+ log ⁡x )^2 Now, ( dy dx )_x=1= log ⁡1 (1+ log ⁡1 )^2=0

$If x^{y} = e^{x - y}, then dy / dx at x = 1 is$

Solution

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