let ( f ) be the function given by ( f(x)=2^{x^{2}} ). selected values of ( f ) are given in the table…

let ( f ) be the function given by ( f(x)=2^{x^{2}} ). selected values of ( f ) are given in the table above. if the values in the table are used to approximate ( f^{prime}(0.5) ), what is the difference between the approximation and the actual value of ( f^{prime}(0.5) )?

let ( f ) be the function given by ( f(x)=2^{x^{2}} ). selected values of ( f ) are given in the table above. if the values in the table are used to approximate ( f^{prime}(0.5) ), what is the difference between the approximation and the actual value of ( f^{prime}(0.5) )?

Answer

Explanation:

Step1: Use the difference quotient for approximation

The difference quotient formula is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b=1$. So the approximation of $f^{\prime}(0.5)$ is $\frac{f(1)-f(0)}{1 - 0}$. Substitute $f(0)=1$ and $f(1)=2$ into the formula: $\frac{2 - 1}{1-0}=1$.

Step2: Find the actual derivative

Given $y = f(x)=2^{x^{2}}$. Use the chain - rule. If $y = a^{u}$ ($a>0,a\neq1$), then $y^{\prime}=a^{u}\ln a\cdot u^{\prime}$. Let $u = x^{2}$, then $u^{\prime}=2x$. So $f^{\prime}(x)=2^{x^{2}}\ln 2\cdot(2x)$. Substitute $x = 0.5$ into $f^{\prime}(x)$: $f^{\prime}(0.5)=2^{(0.5)^{2}}\ln 2\cdot(2\times0.5)$. Since $2^{(0.5)^{2}}=2^{0.25}=\sqrt[4]{2}\approx1.1892$, then $f^{\prime}(0.5)\approx1.1892\times0.6931\times1\approx0.813$.

Step3: Calculate the difference

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