Problems on calculus involving trigonometric function?

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You're off by a sign.

Let y = arcsin((1 - x²)/(1 + x²)), so that sin(y) = (1 - x²)/(1 + x²) = 2/(1 + x²) - 1.

Using a representative triangle, it is easy to conclude that

sec(y) = (1 + x²)/(2x).

Using implicit differentiation

cos(y) dy/dx = -4x/(1 + x²) ==> dy/dx = -4x/(1 + x²) sec(y).

Substituting the known form of sec(y) gives

dy/dx = -2/(1 + x²).

(Work smart----not hard!)

in case you need to use the L'Hopital Rule, this is elementary. merely persist with the rule two times to get: lim (x^2)/(a million - cos x) = lim (2x)/(sin x) = lim 2/(cos x) with all limits as x->0. the main suitable expression is two, and that's your cut back. without L'Hopital, you may multiply with the aid of (a million + cos x)/(a million + cos x), it is valid in a area (-pi < x < pi) around yet no longer alongside with x=0. So: lim (x^2)/(a million - cos x) = lim (a million + cos x)*(x^2)/(a million - (cos x)^2) = lim (a million + cos x)*(x^2)/(sin x)^2 = lim (a million + cos x)*(x / sin x)^2 back, all limits as x->0. It facilitates to be attentive to (or derive) that lim x/sin x = a million/(lim (sin x)/x) = a million/a million = a million. The information that lim[x->0] of x/(sin x) is a million could be recent in maximum calculus books in the section the place derivatives of the trig applications are derived.