If f(0) = f(1) = 0, f’(1) = 2 and y = f(ex) ef(x), write the value of 
at x = 0.
Using the Chain Rule of Differentiation,
= f(ex). ef(x) f’(x) + f’(ex)ex. ef(x)
At x = 0,
= f(1). ef(0) f’(0) + f’(1). ef(0)
= 0. e0 f’(0) + 2.e0
= 0 + 2.1
= 2
1
If f(0) = f(1) = 0, f’(1) = 2 and y = f(ex) ef(x), write the value of at x = 0.
Using the Chain Rule of Differentiation,
= f(ex). ef(x) f’(x) + f’(ex)ex. ef(x)
At x = 0,
= f(1). ef(0) f’(0) + f’(1). ef(0)
= 0. e0 f’(0) + 2.e0
= 0 + 2.1
= 2