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If $f(x)=e^x g(x), g(0)=2, g'(0)=1$ then $f'(0)$ is equal to : |
0 1 2 3 |
3 |
$f(x) = e^xg(x)~~~g(0)=2~~~g'(0)=1$ so $f(0) = e^0(g(0))$ f(0) = 2 differentiating f(x) wrt x using product rule so $f'(x) = \frac{d}{dx} (e^xg(x))$ so $f'(x) = e^xg'(x) + e^xg(x)$ so f'(0) = e0g'(0) + e0g(0) = 2 + 1 = 3 |
