Suppose a person wants to travel D miles at a constant speed of $(10+x) \mathrm{mi} / \mathrm{hr}$, where $x$ could be positive or negative The time in minutes required to travel $D$ miles is $T(x)=60 D(10+x)^{-1}$. Show that the linear approximation to $T$ at the point $x=0$ is $T(x) \approx L(x)=3 D\left(2-\frac{x}{5}\right)$.
Recall that the linear approximation $L(x)$ is equal to $T(a)+T^{\prime}(a)(x-a)$. Find $T^{\prime}(x)$.
\[
T^{\prime}(x)=\square
\]
Answer
The derivative of \(T(x)\) is: \(\boxed{T'(x) = -\frac{60D}{(10+x)^2}}\)
Step 1 ：Rewrite the function as: \(T(x) = 60D(10+x)^{-1}\)
Step 2 ：Apply the power rule to find the derivative of \((10+x)^{-1}\): \(\frac{d}{dx}[(10+x)^{-1}] = -(10+x)^{-2}\)
Step 3 ：Apply the chain rule to find the derivative of \(T(x)\): \(T'(x) = 60D \cdot (-1) \cdot (10+x)^{-2} \cdot 1\)
Step 4 ：Simplify the expression: \(T'(x) = -\frac{60D}{(10+x)^2}\)
Step 5 ：The derivative of \(T(x)\) is: \(\boxed{T'(x) = -\frac{60D}{(10+x)^2}}\)
