Ugly Derivative

Calculus Level 1

$\large y = \cos^{-1}\cos(\log_{2}2^{\ln e^{\sin^{-1}\sin x}})$

For $y$ as defined above, find $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{4}$ .

The answer is 1.

2 solutions

Chew-Seong Cheong
Oct 1, 2015

$\begin{aligned} y & = \cos^{-1} \cos{\left(\log_2 2^{\ln e^{\color{#3D99F6}\sin^{-1} \sin{x}}} \right)} & \small \color{#3D99F6} \text{for } - \frac \pi 2 \le x \le \frac \pi 2 \\ & = \cos^{-1} \cos{\left(\log_2 2^{\ln e^{\color{#3D99F6}x}} \right)} \\ & = \cos^{-1} \cos{\left(\log_2 2^{x} \right)} \\ & = \color{#3D99F6} \cos^{-1} \cos{x} & \small \color{#3D99F6} \text{for }0 \le x \le \pi \\ & =\color{#3D99F6} x \end{aligned}$

$\implies \dfrac{dy}{dx} = \boxed{1}$

Note that this only works for $x \in [0, \frac{\pi}{2}]$ due to the constraints on the ranges of $\sin^{-1} x$ and $\cos^{-1} x.$

Steven Yuan - 3 years, 2 months ago

can anyone please explain how the equation was symplified to x?! i tried applying the chain rule and got a mess.

j parra - 3 years, 2 months ago

inverse function

David Ding - 3 years, 2 months ago

In general, $y = f(x)$ , $\implies x = f^{-1} (y)$ . $y= \cos x$ , $\implies \cos^{-1} (y) = x$ , $\implies \cos^{-1} (\cos x) = x$ .

Chew-Seong Cheong - 3 years, 2 months ago

Deepak Kumar
Sep 30, 2015

By observation,y=x=>dy/dx=1

Ugly Derivative

The answer is 1.

2 solutions

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