How do you find the derivative of $(sqrt(1-x^2))arcsinx$ ?

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Answer 1 · 2017-09-07T10:24:11

$1-(x*arc sinx)/sqrt(1-x^2), |x|<1.$

Let, $y=sqrt(1-x^2)*arc sinx.$

The desired Derivative can be obtained using the Product Rule for

Diffn.

Here is, an another way to solve the Problem.

We subst. $t=arc sinx, |x|<=1. rArr sint=x, |t| <=pi/2.$

Hence, $y=sqrt(1-sin^2t)*t=tcost," where, "t=sinx....(1).$

Thus, $y" is a fun. of "t," and, t of "x.$

By the Chain Rule, then, we have,

$dy/dx=dy/dt*dt/dx,$

$={d/dt(tcost)}{d/dx(arc sinx)}......[because, (1)],$

$={cost*d/dt(t)+t*d/dt(cost)}(1/sqrt(1-x^2)),$

$={cost+t(-sint)}cosx,$

$={sqrt(1-x^2)-x*arc sinx}(1/sqrt(1-x^2)),$

$rArr dy/dx=1-(x*arc sinx)/sqrt(1-x^2), |x|<1.$

N.B. : The Domain of $g" is "[-1,1];" and, that of "g'" is "(-1,1).$

Enjoy Maths.!

How do you find the derivative of (sqrt(1-x^2))arcsinx(sqrt(1-x^2))arcsinx?