If f (1)=3, f ^ (1)=2 , then d d x [ f ( e ^x+2 x ) ] at x=0 is

Step-by-step solution

Given y = [f (e^x + 2x ) ] , the derivative dy dx is found using the chain rule. Let u = f(e^x + 2x) , so y = u and dy du = 1 u . Differentiating u with respect to v = e^x + 2x gives du dv = f'(v) , while dv dx = e^x + 2 . Combining these, dy dx = 1 f(e^x + 2x) f'(e^x + 2x) (e^x + 2) . At x = 0 , e^0 + 2 0 = 1 . Substituting f(1) = 3 and f'(1) = 2 : . dy dx |_ x=0 = 1 3 2 (1 + 2) = 2 . The derivative at x = 0 is 2 .