If f(x) = b e^{ax} + a e^{bx},then f^{prime p | vedclass

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(C) Given,$f(x) = b e^{ax} + a e^{bx}$.First,differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}(b e^{ax} + a e^{bx}) = b(a e^{ax})

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$ab(a + b)$

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(C) Given,$f(x) = b e^{ax} + a e^{bx}$.First,differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}(b e^{ax} + a e^{bx}) = b(a e^{ax}) + a(b e^{bx}) = ab e^{ax} + ab e^{bx}$.Next,differentiate $f^{\prime}(x)$ with respect to $x$ to find the second derivative:$f^{\prime \prime}(x) = \frac{d}{dx}(ab e^{ax} + ab e^{bx}) = ab(a e^{ax}) + ab(b e^{bx}) = a^2 b e^{ax} + ab^2 e^{bx}$.Finally,evaluate at $x = 0$:$f^{\prime \prime}(0) = a^2 b e^{a(0)} + ab^2 e^{b(0)} = a^2 b(1) + ab^2(1) = a^2 b + ab^2$.Factoring out $ab$,we get:$f^{\prime \prime}(0) = ab(a + b)$.

If $f(x) = b e^{ax} + a e^{bx}$,then $f^{\prime \prime}(0)$ is equal to

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