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.!