So much inverse

Calculus Level 3

f ( x ) = arctan ⁡ [ 1 + sin ⁡ x + 1 − sin ⁡ x 1 + sin ⁡ x − 1 − sin ⁡ x ] f\left( x \right) =\arctan { \left[\, \frac { \sqrt { 1+\sin { x } } +\sqrt { 1-\sin { x } } }{ \sqrt { 1+\sin { x } } -\sqrt { 1-\sin { x } } } \, \right] }

Find the derivative of f ( x ) f(x) at x = π 4 . x = \frac\pi4.


The answer is -0.5.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

1 + sin ⁡ ( 2 β ) = ( sin ⁡ β + cos ⁡ β ) 2 = sin ⁡ β + cos ⁡ β {\color{#20A900}{\sqrt{1 + \sin(2β)} = \sqrt{(\sin β + \cos β)^2} = \sin β + \cos β}}

f ( x ) = arctan ⁡ [ ( sin ⁡ x / 2 + cos ⁡ x / 2 ) + ( sin ⁡ x / 2 − cos ⁡ x / 2 ) ( sin ⁡ x / 2 + cos ⁡ x / 2 ) − ( sin ⁡ x / 2 − cos ⁡ x / 2 ) ] f(x) = \arctan \begin{bmatrix} \dfrac {(\sin x/2 + \cos x/2 )+ ( \sin x/2 - \cos x/2)}{(\sin x/2 + \cos x/2) - (\sin x/2 - \cos x/2)} \end{bmatrix}

f ( x ) = arctan ⁡ [ 2 sin ⁡ x / 2 2 cos ⁡ x / 2 ] f(x) = \arctan \begin{bmatrix} \dfrac {2\sin x/2 }{2\cos x/2} \end{bmatrix}

f ( x ) = arctan ⁡ [ tan ⁡ ( x 2 ) ] f(x) = \arctan \begin{bmatrix} \tan \begin{pmatrix} \dfrac x2 \end{pmatrix} \end{bmatrix}

f ( x ) = x 2 ⇒ f ′ ( x ) = 1 2 f(x) = \dfrac x2 \Rightarrow f'(x) = \dfrac 12

Hence, f ( π 4 ) = 1 2 = A B f\begin{pmatrix} \dfrac π4 \end{pmatrix} = \dfrac 12 = \dfrac AB

So, A + B = 3 A + B = \boxed{3}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

(January 26 2021)

This solution is currently wrong.

Well there is a error here... we require the derivative at pi/4. For x belonging to (0,pi/2) ,cos(x/2) > sin(x/2), so the expression for f(x) wont be x/2 since sqrt(x^2) = modulus(x). What we will obtain is arctan(cot(x/2)) (here) .In the end, we get f'(pi/4) = -1/2 not 1/2 as f(x) will be pi/2 - x/2

Natalia Dcruz - 2 years, 3 months ago

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