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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 $$a=1$$

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

$$x=x \quad a=1$$

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

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

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




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


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