Suppose, f(x)=e^{-sqrt{x}}+e^{-frac{1}{x^2}}. | vedclass

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(B) Given the function $f(x)=e^{-\sqrt{x}}+e^{-\frac{1}{x^2}}$.First, we find the first derivative $f^{\prime}(x)$ using the chain rule:$f^{\prime}(x)

Vedclass · Accepted Answer

$\left(\frac{1}{4},-2\right)$

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(B) Given the function $f(x)=e^{-\sqrt{x}}+e^{-\frac{1}{x^2}}$.First, we find the first derivative $f^{\prime}(x)$ using the chain rule:$f^{\prime}(x) = e^{-\sqrt{x}} \cdot \frac{d}{dx}(-\sqrt{x}) + e^{-\frac{1}{x^2}} \cdot \frac{d}{dx}(-x^{-2})$$f^{\prime}(x) = e^{-\sqrt{x}} \left(-\frac{1}{2\sqrt{x}}\right) + e^{-\frac{1}{x^2}} \left(\frac{2}{x^3}\right)$Now, differentiate $f^{\prime}(x)$ again to find $f^{\prime \prime}(x)$:$f^{\prime \prime}(x) = \frac{d}{dx} \left[ -\frac{1}{2} x^{-1/2} e^{-\sqrt{x}} \right] + \frac{d}{dx} \left[ 2 x^{-3} e^{-\frac{1}{x^2}} \right]$Using the product rule:For the first term: $-\frac{1}{2} \left[ e^{-\sqrt{x}} \cdot (-\frac{1}{2} x^{-3/2}) + x^{-1/2} \cdot e^{-\sqrt{x}} \cdot (-\frac{1}{2} x^{-1/2}) \right] = \frac{1}{4} \frac{e^{-\sqrt{x}}}{x} (1 + \frac{1}{\sqrt{x}})$For the second term: $2 \left[ e^{-\frac{1}{x^2}} \cdot (-3x^{-4}) + x^{-3} \cdot e^{-\frac{1}{x^2}} \cdot (2x^{-3}) \right] = 2 \frac{e^{-\frac{1}{x^2}}}{x^4} (-3 + \frac{2}{x^2}) = -2 \frac{e^{-\frac{1}{x^2}}}{x^4} (3 - \frac{2}{x^2})$Comparing this with the given expression, we get $\alpha = \frac{1}{4}$ and $\beta = -2$.

Suppose, $f(x)=e^{-\sqrt{x}}+e^{-\frac{1}{x^2}}$. If $f^{\prime \prime}(x)=\alpha \cdot \frac{e^{-\sqrt{x}}}{x}\left(1+\frac{1}{\sqrt{x}}\right)+\beta \cdot \frac{e^{-\frac{1}{x^2}}}{x^4}\left(3-\frac{2}{x^2}\right)$, then $(\alpha, \beta)=$

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