# How do you find the derivative of f(x)=sqrt(x+1)?

##### 1 Answer
Jul 9, 2016

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

#### Explanation:

$f \left(x\right) = \sqrt{x + 1} = {\left(x + 1\right)}^{\frac{1}{2}} ,$ and, let $\left(x + 1\right) = t , \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$ so that,

$f \left(x\right) = {t}^{\frac{1}{2}} , \mathmr{and} , t = x + 1.$

This means that, $f$ is a fun. of $t$, and, $t ,$ in turn, that of $x$.

In such cases, the desired Diffn. has to be carried out by using a Rule, called The Chain Rule given by,

$f ' \left(x\right) = \frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} \ldots \ldots \ldots . . \left(2\right)$

In $\left(2\right)$, we need $\frac{\mathrm{df}}{\mathrm{dt}}$ & $\frac{\mathrm{dt}}{\mathrm{dx}}$, which we work out below :-

$\frac{\mathrm{df}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({t}^{\frac{1}{2}}\right) \Rightarrow \frac{1}{2} \cdot {t}^{\frac{1}{2} - 1} = \frac{1}{2} \cdot {t}^{- \frac{1}{2}} \ldots \ldots . . \left(3\right) .$

Also, $\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x + 1\right) = \frac{d}{\mathrm{dx}} x + \frac{d}{\mathrm{dx}} 1 = 1 + 0 = 1. \ldots \ldots \ldots \ldots . \left(4\right)$

Using $\left(2\right) , \left(3\right) , \left(4\right)$ we have,

$f ' \left(x\right) = \frac{1}{2} \cdot {t}^{- \frac{1}{2}} \cdot 1 = \frac{1}{2 {t}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{t}} = \frac{1}{2 \sqrt{x + 1}} \ldots \ldots \ldots . . \left[b y \left(1\right)\right] .$

Hope, you enjoyed it!