If f(x)=b cdot e^{a x}+a cdot e^{b x},then f^ | vedclass

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(D) Given the function $f(x) = b \cdot e^{ax} + a \cdot e^{bx}$.First,find the first derivative $f^{\prime}(x)$ with respect to $x$:$f^{\prime}(x) = \

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$a b(a+b)$

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

If $f(x)=b \cdot e^{a x}+a \cdot e^{b x}$,then $f^{\prime \prime}(0)=$

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