determine $f^{(4)}(x)$ for $f(x)=0.125x^{6}-\frac{1}{x}+e^{-\frac{x}{2}} - 55.$

determine $f^{(4)}(x)$ for $f(x)=0.125x^{6}-\frac{1}{x}+e^{-\frac{x}{2}} - 55.$

determine $f^{(4)}(x)$ for $f(x)=0.125x^{6}-\frac{1}{x}+e^{-\frac{x}{2}} - 55.$

Answer

Explanation:

Step1: Recall power - rule and chain - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$, and for a composite function $y = f(g(x))$, the chain - rule states that $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Also, the derivative of $e^{u}$ with respect to $x$ is $e^{u}\cdot u^\prime$, where $u$ is a function of $x$.

Step2: Differentiate each term of $f(x)=0.125x^{6}-\frac{1}{x}+e^{-\frac{x}{2}} - 55$

  1. For the first term $y_1 = 0.125x^{6}$, using the power - rule, $y_1^\prime=0.125\times6x^{5}=0.75x^{5}$.
  2. For the second term $y_2=-\frac{1}{x}=-x^{-1}$, using the power - rule, $y_2^\prime=-(-1)x^{-2}=x^{-2}=\frac{1}{x^{2}}$.
  3. For the third term $y_3 = e^{-\frac{x}{2}}$, using the chain - rule with $u =-\frac{x}{2}$, $y_3^\prime=e^{-\frac{x}{2}}\times(-\frac{1}{2})=-\frac{1}{2}e^{-\frac{x}{2}}$.
  4. For the fourth term $y_4=-55$, since the derivative of a constant is 0, $y_4^\prime = 0$. So, $f^\prime(x)=0.75x^{5}+\frac{1}{x^{2}}-\frac{1}{2}e^{-\frac{x}{2}}$.

Step3: Differentiate $f^\prime(x)$ to get $f^{\prime\prime}(x)$

  1. For the first term $y_1 = 0.75x^{5}$, using the power - rule, $y_1^\prime=0.75\times5x^{4}=3.75x^{4}$.
  2. For the second term $y_2=\frac{1}{x^{2}}=x^{-2}$, using the power - rule, $y_2^\prime=-2x^{-3}=-\frac{2}{x^{3}}$.
  3. For the third term $y_3 =-\frac{1}{2}e^{-\frac{x}{2}}$, using the chain - rule, $y_3^\prime=-\frac{1}{2}e^{-\frac{x}{2}}\times(-\frac{1}{2})=\frac{1}{4}e^{-\frac{x}{2}}$. So, $f^{\prime\prime}(x)=3.75x^{4}-\frac{2}{x^{3}}+\frac{1}{4}e^{-\frac{x}{2}}$.

Step4: Differentiate $f^{\prime\prime}(x)$ to get $f^{(3)}(x)$

  1. For the first term $y_1 = 3.75x^{4}$, using the power - rule, $y_1^\prime=3.75\times4x^{3}=15x^{3}$.
  2. For the second term $y_2=-\frac{2}{x^{3}}=-2x^{-3}$, using the power - rule, $y_2^\prime=-2\times(-3)x^{-4}=\frac{6}{x^{4}}$.
  3. For the third term $y_3=\frac{1}{4}e^{-\frac{x}{2}}$, using the chain - rule, $y_3^\prime=\frac{1}{4}e^{-\frac{x}{2}}\times(-\frac{1}{2})=-\frac{1}{8}e^{-\frac{x}{2}}$. So, $f^{(3)}(x)=15x^{3}+\frac{6}{x^{4}}-\frac{1}{8}e^{-\frac{x}{2}}$.

Step5: Differentiate $f^{(3)}(x)$ to get $f^{(4)}(x)$

  1. For the first term $y_1 = 15x^{3}$, using the power - rule, $y_1^\prime=15\times3x^{2}=45x^{2}$.
  2. For the second term $y_2=\frac{6}{x^{4}}=6x^{-4}$, using the power - rule, $y_2^\prime=6\times(-4)x^{-5}=-\frac{24}{x^{5}}$.
  3. For the third term $y_3=-\frac{1}{8}e^{-\frac{x}{2}}$, using the chain - rule, $y_3^\prime=-\frac{1}{8}e^{-\frac{x}{2}}\times(-\frac{1}{2})=\frac{1}{16}e^{-\frac{x}{2}}$.

Answer:

$f^{(4)}(x)=45x^{2}-\frac{24}{x^{5}}+\frac{1}{16}e^{-\frac{x}{2}}$