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Find the approximate value for $\sqrt[3]{63.9}$

(1) $x=63.9$

(2) $a=64$

(3) $f(x)=\sqrt[3]{x}=x^{1 / 3}$

(4) $f(a)=f(64)=\sqrt[3]{64}=4$

(5) $f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}$

(6) $f^{\prime}(a)=f^{\prime}(64)=\frac{1}{3}(64)^{-2 / 3}=0.02=\frac{1}{48}$

(7) $f(x) \simeq L(x)=f(a)+f^{\prime}(a)(x-a)$

(8) $f(63.9) \simeq L(63.9)=4+\frac{1}{48}(63.9-64)$

$f(63.9) \simeq 4+\frac{1}{48}\left(-\frac{1}{10}\right)$

$f(63.9) \simeq 4-\frac{1}{480}$

$\sqrt[3]{63.9} \simeq 4-\frac{1}{480}$

Find the differential of the functions:

$y=\frac{s}{1+3 s}$
$y=\tan \sqrt{t}$
$y=x^{2} \sin 2 x$
$y=\ln \sqrt{1+t^{2}}$

$y=\frac{s}{1+3 s} \longrightarrow d y=f^{\prime}(s) d s$

$\rightarrow d y=\left(\frac{(1)(1+3 s)-(3)(s)}{(1+3 s)^{2}}\right) d s$

$d y=\frac{1}{(1+3s)^{2}} d s$

$y=\tan \sqrt{t} \rightarrow d y=f ^\prime(t) d t$

$\rightarrow d y=\left(\sec ^{2} \sqrt{t} \cdot \frac{1}{2 \sqrt{t}}\right) d t$

$y=x^{2} \sin 2 x \rightarrow d y=f^{\prime}(x) d x$

$\rightarrow d y=\left(2 x \sin 2 x+2 x^{2} \cos 2 x\right) d x$

$y=\ln \sqrt{1+t^{2}} \rightarrow d y=f^{\prime}(t) d t$

$\rightarrow d y=\left(\frac{2t/ 2 \sqrt{1+t^2}}{\sqrt{1+t^{2}}}\right) d t=\left(\frac{t}{1+t^{2}}\right) d t$

Find the linear approximation of $f(x)=\sqrt{2 x+7}$ at $a=1$

$f(x)=\sqrt{2 x+7} \quad, \quad a=1$

$x=x \quad a=1$

$f(x)=\sqrt{2 x+7}$

$f^{\prime}(x)=\frac{2}{2 \sqrt{2 x+7}}=\frac{1}{\sqrt{2 x+7}}$

$f(x) \simeq L(x)=f(a)+f^{\prime}(a)(x-a)$

$a=1$

$f(a)=f(1)=\sqrt{2(1)+7}=\sqrt{9}=3$

$f^{\prime}(a)=\frac{1}{\sqrt{2(1)+7}}=\frac{1}{\sqrt{9}}=\frac{1}{3}$

$f(x) \simeq L(x)=f(a)+f^{\prime}(a)(x-a)$

$=3+\frac{1}{3}(x-1)$