## University Calculus: Early Transcendentals (3rd Edition)

$$\left.\frac{df^{-1}}{dx}\right|_{x=4}=3$$
Here we do not know the formula for the function $y=f(x)$. But we know that it is differentiable, it has an inverse and its graph passes through the point $(2,4)$ and has a slope of $1/3$ there. Since the value of $df/dx$ at a point $x=a$ is also the value of the slope at that point, the given information tells us that $$\left.\frac{df}{dx}\right|_{x=2}=\frac{1}{3}$$ Now we need to find $df^{-1}/dx$ at $x=4=f(2)$. According to Theorem 3, $$\left.\frac{df^{-1}}{dx}\right|_{x=4=f(2)}=\frac{1}{\left.\frac{df}{dx}\right|_{x=2}}=\frac{1}{\frac{1}{3}}=3$$