What is the derivative of ${arctan (1 + 2 x)}^{\frac{1}{2}}$ $arctan(1+2x)^(1/2)$ ?

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Answer 1 · 2016-07-13T10:43:52

$\frac{d y}{d x} = \frac{1}{2 (1 + x) \sqrt{1 + 2 x}} .$ $dy/dx=1/{2(1+x)sqrt(1+2x)}.$

Let $y = {arctan (1 + 2 x)}^{\frac{1}{2}}$ $y=arctan(1+2x)^(1/2)$

Put $u = {(1 + 2 x)}^{\frac{1}{2}}$ $u=(1+2x)^(1/2)$ & $(1 + 2 x) = t .$ $(1+2x)=t.$

With these substitutions, we see that, $y=arctanu, u=t^(1/2), &, t=1+2x...............(1).$

Thus, $y$ is a function of $u$ , $u$ of $t$ , &, $t$ of $x$ .

Accordingly, to find $dy/dx$ we need to apply the Chain Rule, which states that in the situation described above,

$dy/dx=dy/(du)*(du)/(dt)*dt/dx$

$={d/(du)(arctanu))}{d/dt(t^(1/2)}{d/dx(1+2x)}$

$={1/(1+u^2)}{(1/2)*t^(-1/2)}(0+2)$

$=1/{1+((1+2x)^(1/2))^2}{1/t^(1/2)}$

$={1/(1+1+2x)}*{1/(1+2x)^(1/2)}$

$=1/(2(1+x))*1/(1+2x)^(1/2)$

$:. dy/dx=1/{2(1+x)sqrt(1+2x)}.$

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