How do you find the tangent line approximation for $f (x) = \sqrt{1 + x}$ $f(x)=sqrt(1+x)$ near $x = 0$ $x=0$ ?

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Answer 1 · 2014-09-24T04:11:03

We need to find the derivative of $f (x)$ $f(x)$ . We need to use the Chain Rule to find the derivative of $f (x)$ $f(x)$ .

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

$f'(x)=(1/2)(1+x)^((1/2-1))*1$

$f'(x)=(1/2)(1+x)^((1/2-2/2))$

$f'(x)=(1/2)(1+x)^((-1/2))$

$f'(x)=1/(2sqrt(1+x))$

$f'(0)=1/(2sqrt(1+0))=1/(2sqrt(1))=1/2=0.5$

How do you find the tangent line approximation for f(x)=sqrt(1+x)f(x)=√1+xf(x)=sqrt(1+x) near x=0x=0x=0 ?