Question

1. - / 1 points find the derivative of the function. y = tan^(-1)(sqrt(x - 4)) y =

Explanation:

Step1: Recall the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = \sqrt{x - 4}$, so $y=\tan^{- 1}(u)$.

Step2: Find the derivative of $y$ with respect to $u$

The derivative of $y = \tan^{-1}(u)$ with respect to $u$ is $y'_u=\frac{1}{1 + u^{2}}$. Since $u=\sqrt{x - 4}$, we have $y'_u=\frac{1}{1+(\sqrt{x - 4})^{2}}=\frac{1}{1+(x - 4)}=\frac{1}{x - 3}$.

Step3: Find the derivative of $u$ with respect to $x$

Since $u=(x - 4)^{\frac{1}{2}}$, using the power - rule $(x^n)'=nx^{n - 1}$, we get $u'_x=\frac{1}{2}(x - 4)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x - 4}}$.

Step4: Apply the chain - rule

By the chain - rule $y'=y'_u\cdot u'_x$. Substituting the values of $y'_u$ and $u'_x$ we found above, we have $y'=\frac{1}{2\sqrt{x - 4}(x - 3)}$.

Answer:

$\frac{1}{2\sqrt{x - 4}(x - 3)}$