Question

determine the derivative of each of the following functions.
a. $f(x)=\frac{1}{7}\tan^{-1}(\frac{x}{7})$
$f(x)=$
b. $g(x)=\frac{x}{2}sqrt{49 - x^{2}}+24.5sin^{-1}(\frac{x}{7})$
$g(x)=$

Explanation:

Step1: Recall derivative of inverse - tangent function

The derivative of $y = \tan^{-1}(u)$ is $y'=\frac{u'}{1 + u^{2}}$. For $f(x)=\frac{1}{7}\tan^{-1}(\frac{x}{7})$, let $u = \frac{x}{7}$, then $u'=\frac{1}{7}$.
$f'(x)=\frac{1}{7}\times\frac{\frac{1}{7}}{1+(\frac{x}{7})^{2}}$

Step2: Simplify the expression

$f'(x)=\frac{1}{7}\times\frac{\frac{1}{7}}{1+\frac{x^{2}}{49}}=\frac{1}{7}\times\frac{\frac{1}{7}}{\frac{49 + x^{2}}{49}}=\frac{1}{49 + x^{2}}$

Step3: For $g(x)=\frac{x}{2}\sqrt{49 - x^{2}}+24.5\sin^{-1}(\frac{x}{7})$

Use the product - rule $(uv)' = u'v+uv'$ for the first term $\frac{x}{2}\sqrt{49 - x^{2}}$, where $u=\frac{x}{2}$ and $v=\sqrt{49 - x^{2}}=(49 - x^{2})^{\frac{1}{2}}$.
$u'=\frac{1}{2}$, and $v'=\frac{1}{2}(49 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{49 - x^{2}}}$.
The derivative of the first term is $\frac{1}{2}\sqrt{49 - x^{2}}+\frac{x}{2}\times\frac{-x}{\sqrt{49 - x^{2}}}=\frac{49 - x^{2}-x^{2}}{2\sqrt{49 - x^{2}}}=\frac{49 - 2x^{2}}{2\sqrt{49 - x^{2}}}$.
The derivative of the second term: The derivative of $y=\sin^{-1}(u)$ is $y'=\frac{u'}{\sqrt{1 - u^{2}}}$. For $y = 24.5\sin^{-1}(\frac{x}{7})$, let $u=\frac{x}{7}$, $u'=\frac{1}{7}$, then its derivative is $24.5\times\frac{\frac{1}{7}}{\sqrt{1 - (\frac{x}{7})^{2}}}=\frac{24.5}{7\sqrt{1-\frac{x^{2}}{49}}}=\frac{24.5}{7}\times\frac{7}{\sqrt{49 - x^{2}}}=\frac{24.5}{\sqrt{49 - x^{2}}}$.
$g'(x)=\frac{49 - 2x^{2}}{2\sqrt{49 - x^{2}}}+\frac{24.5}{\sqrt{49 - x^{2}}}=\frac{49 - 2x^{2}+49}{2\sqrt{49 - x^{2}}}=\frac{98 - 2x^{2}}{2\sqrt{49 - x^{2}}}=\sqrt{49 - x^{2}}$

Answer:

a. $f'(x)=\frac{1}{49 + x^{2}}$
b. $g'(x)=\sqrt{49 - x^{2}}$