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If $f(x)=b e^{a x}+a e^{b x}$, then $f^{\prime \prime}(0)$ is equal to
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Verified Answer
The correct answer is:
$a b(a+b)$
Given,
$$
f(x)=b e^{a x}+a e^{b x}
$$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=a b e^{a x}+a b e^{b x}
$$
Again, differentiating, we get
$$
$$
\begin{aligned}
\mathrm{f}^{\prime \prime}(\mathrm{x}) &=\mathrm{a}^{2} b \mathrm{e}^{\mathrm{ax}}+\mathrm{ab}^{2} \mathrm{e}^{\mathrm{bx}} \\
\Rightarrow \quad \mathrm{f}^{\prime \prime}(0) &=\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2} \\
&=\mathrm{ab}(\mathrm{a}+\mathrm{b})
\end{aligned}
$$
$$
f(x)=b e^{a x}+a e^{b x}
$$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=a b e^{a x}+a b e^{b x}
$$
Again, differentiating, we get
$$
$$
\begin{aligned}
\mathrm{f}^{\prime \prime}(\mathrm{x}) &=\mathrm{a}^{2} b \mathrm{e}^{\mathrm{ax}}+\mathrm{ab}^{2} \mathrm{e}^{\mathrm{bx}} \\
\Rightarrow \quad \mathrm{f}^{\prime \prime}(0) &=\mathrm{a}^{2} \mathrm{~b}+\mathrm{ab}^{2} \\
&=\mathrm{ab}(\mathrm{a}+\mathrm{b})
\end{aligned}
$$
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