suppose the sun is shining on a pole that is 4 feet tall. if the pole casts a shadow that is $x$ feet long…

Explanation:

Step1: Differentiate $\theta$ with respect to $x$

We know that $\theta=\arctan\left(\frac{4}{x}\right)$. Using the chain - rule and the derivative formula for $\arctan(u)$ which is $\frac{d}{du}\arctan(u)=\frac{1}{1 + u^{2}}$. Let $u = \frac{4}{x}$, then $\frac{d\theta}{dx}=\frac{1}{1+\left(\frac{4}{x}\right)^{2}}\cdot\frac{d}{dx}\left(\frac{4}{x}\right)$.

Step2: Differentiate $\frac{4}{x}$

Since $\frac{4}{x}=4x^{-1}$, then $\frac{d}{dx}(4x^{-1})=- 4x^{-2}=-\frac{4}{x^{2}}$.

Step3: Simplify $\frac{d\theta}{dx}$

Substitute $\frac{d}{dx}\left(\frac{4}{x}\right)=-\frac{4}{x^{2}}$ into the expression for $\frac{d\theta}{dx}$. We have $\frac{d\theta}{dx}=\frac{1}{1 + \frac{16}{x^{2}}}\cdot\left(-\frac{4}{x^{2}}\right)$. First, simplify $1+\frac{16}{x^{2}}=\frac{x^{2}+16}{x^{2}}$. Then $\frac{d\theta}{dx}=\frac{x^{2}}{x^{2}+16}\cdot\left(-\frac{4}{x^{2}}\right)=-\frac{4}{x^{2}+16}$.

Step4: Evaluate at $x = 6$

When $x = 6$, $\frac{d\theta}{dx}=-\frac{4}{6^{2}+16}=-\frac{4}{36 + 16}=-\frac{4}{52}=-\frac{1}{13}$ radians per foot.

Answer:

$-\frac{1}{13}$