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

Sep 24, 2014

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

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

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(\frac{1}{2} - 1\right)} \cdot 1$

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(\frac{1}{2} - \frac{2}{2}\right)}$

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(1 + x\right)}^{\left(- \frac{1}{2}\right)}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{1 + x}}$

$f ' \left(0\right) = \frac{1}{2 \sqrt{1 + 0}} = \frac{1}{2 \sqrt{1}} = \frac{1}{2} = 0.5$