Linear Approximation and Differentials
$$dy = f'(x)\, dx$$
$$\Delta y \approx dy = f'(x)\Delta x$$
$$L(x) = f(a) + f'(a)(x-a)$$
Find $dy$ if $y = x^3 - 4x + 1$.
The differential is $dy = f'(x)\,dx$.
Use differentials to approximate the change in $y = \sqrt{x}$ as $x$ increases from 25 to 26.
$f(x) = \sqrt{x}$, $f'(x) = \dfrac{1}{2\sqrt{x}}$, $a = 25$, $dx = 1$.
So $\sqrt{26} \approx 5 + 0.1 = 5.1$. (Actual: $\sqrt{26} \approx 5.0990$)
Approximate $\sqrt[3]{8.1}$ using the linearization $L(x) = f(a) + f'(a)(x - a)$.
Let $f(x) = x^{1/3}$, $a = 8$.
(Actual: $\sqrt[3]{8.1} \approx 2.00829$)
The radius of a sphere is measured as $r = 6$ cm with a possible error of $dr = 0.05$ cm. Estimate the relative percentage error in the surface area $S = 4\pi r^2$.
Relative error:
Find $dy$ if $y = x^2 \sin x$.
Apply the product rule: