Question #b6b90

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Answer 1 · 2017-11-02T23:16:29

y'=arcsenx+ $\frac{x}{\sqrt (1 - x^{2})} + \frac{1}{\sqrt x (2 x + 2)}$ $x/(√(1-x^2))+1/(√x(2x+2))$

Use product rule
y=xarcsenx
y'=x'(arcsenx)+x(arcsenx)'

Derivative of x=1
Derivative of arcsenx= $\frac{1}{\sqrt (1 - x^{2})}$ $1/(√(1-x^2))$
Then
y'=arcsenx+ $\frac{x}{\sqrt (1 - x^{2})}$ $x/(√(1-x^2))$

Now for $a r c t g \sqrt x$ $arctg√x$

derivative of arctg(u) where u is a function

Is $\frac{\frac{d}{d x} (u)}{1 + u^{2}}$ $(d/dx(u))/(1+u^2)$
Then for $\sqrt x$ $√x$

Derivative of √x is $\frac{1}{2 \sqrt x}$ $1/(2√x)$
Then
Derivative of arctg√x= $\frac{\frac{1}{2 \sqrt x}}{1 + x}$ $(1/(2√x))/(1+x)$

-> $\frac{1}{\sqrt x (2 x + 2)}$ $1/(√x(2x+2))$