How do you find the derivative of #y = 9tan^-1(x − (sqrt (1 + x^2))#?

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Answer 1 · 2016-09-05T06:53:51

#dy/dx=9/(2(1+x^2)), x>0#.

We give soln. for #x>0#.

Let, #x=tan theta, x>0", so that, "theta=tan^-1 x, theta in (-pi/2,pi/2)#.

Since, #x>0, theta in (0,pi/2)#.

Now, #(x-sqrt(1+x^2))=(tan theta-sectheta)=(sintheta-1)/costheta#

#={2sin(theta/2)cos(theta/2)-sin^2(theta/2)-cos^2(theta/2)}/(cos^2(theta/2)-sin^2(theta/2)#

#=(cos^2(theta/2)+sin^2(theta/2)-2sin(theta/2)cos(theta/2)}/(sin^2(theta/2)-cos^2(theta/2)#

#=(sin(theta/2)-cos(theta/2))^2/{(sin(theta/2)-cos(theta/2))(sin(theta/2)+cos(theta/2))#

#=(sin(theta/2)-cos(theta/2))/(sin(theta/2)+cos(theta/2))#

#=tan(theta/2-pi/4)#

Observe that both the sides are #-ve#, so the eqn. is OK.

Hence, #y=9tan^-1(x-sqrt(1+x^2))=9tan^-1(tan(theta/2-pi/4))#

#=9(theta/2-pi/4)=9/2(tan^-1 x)-9pi/4#

#:. dy/dx=(9/2)(1/(1+x^2))=9/(2(1+x^2)), x>0#.

The Case : x<0 can be dealt with similarly.

Enjoy maths.!