Q11 of 354 Page 11

If f(0) = f(1) = 0, f’(1) = 2 and y = f(e^x) e^f(x), write the value of at x = 0.

Using the Chain Rule of Differentiation,

= f(e^x). e^f(x) f^’(x) + f^’(e^x)e^x. e^f(x)

At x = 0,

= f(1). e^f(0) f’(0) + f’(1). e^f(0)

= 0. e⁰ f’(0) + 2.e⁰

= 0 + 2.1

= 2

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If f(0) = f(1) = 0, f’(1) = 2 and y = f(ex) ef(x), write the value of at x = 0.