Let h(x) be differentiable for all x and let | vedclass

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(C) Given $f(x) = (kx + e^x) h(x)$.Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(kx + e^x) \cdot h(x) + (kx + e^x) \cdot h'(x)$.$f'

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$3$

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(C) Given $f(x) = (kx + e^x) h(x)$.Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(kx + e^x) \cdot h(x) + (kx + e^x) \cdot h'(x)$.$f'(x) = (k + e^x) h(x) + (kx + e^x) h'(x)$.Now,substitute $x = 0$ into the derivative expression:$f'(0) = (k + e^0) h(0) + (k(0) + e^0) h'(0)$.$f'(0) = (k + 1) h(0) + (0 + 1) h'(0)$.$f'(0) = (k + 1) h(0) + h'(0)$.Given $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,we substitute these values:$18 = (k + 1)(5) + (-2)$.$18 = 5k + 5 - 2$.$18 = 5k + 3$.$15 = 5k$.$k = 3$.

Let $h(x)$ be differentiable for all $x$ and let $f(x) = (kx + e^x) h(x)$ where $k$ is some constant. If $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,then the value of $k$ is equal to

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