$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:
A$\dfrac{2}{\sqrt{1-x^2}}$
B$\dfrac{-2}{1+x^2}$
C$\dfrac{1}{1+x^2}$
D$\dfrac{2}{1+x^2}$
Answer & Solution
Correct answer: D. $\dfrac{2}{1+x^2}$
Substitute $x = \tan\theta$. Then $(1-x^2)/(1+x^2) = \cos 2\theta$. So $\cos^{-1}\cos 2\theta = 2\theta = 2\tan^{-1}x$ (for $x > 0$, i.e. $0 < \theta < \pi/2$, so $0 < 2\theta < \pi$ — valid range of $\cos^{-1}$). Derivative = $2/(1+x^2)$.
Related questions
An air pollution index is $p=x^2+2xy+4xy^2$. At the point $(10,5)$, the partial derivativeUsing the second derivative test on $y=3xe^{-x}$, the stationary point is:If $y=\tan^{-1}\left(\dfrac{ qrt{1+x}- qrt{1-x}}{ qrt{1+x}+ qrt{1-x}}\right)$, then $\dfraIf $f(x) = e^{ax} in(bx)$, then $f''(x) + a^2 f(x)$ equals:If $f(x) = x^3 - 3x$, the values of $x$ where $f'(x) = 0$ are:If $y_n$ denotes the $n$-th derivative of $y = in x$, then $y_n = $:If $y = \tan^{-1}\left(\dfrac{ qrt{1+x^2} - 1}{x}\right)$, then $\dfrac{dy}{dx}$ equals:If $f(x) = |x - 1| + |x - 3|$, then $f'(2)$ equals: