Tag: Derivatives

Find the derivative of arcsin(sin x)

by RoRi

November 29, 2015

Find the derivative of the function

$f(x) = \arcsin (\sin x).$

Using the chain rule and the formula for the derivative of arcsine we have

$\begin{align*} f'(x) &= (\cos x) \left( \frac{1}{\sqrt{1-\sin^2 x}} \right) \\ &= \frac{\cos x}{\sqrt{\cos^2 x}} \\ &= \frac{\cos x}{| \cos x|}. \end{align*}$

This is valid for all $x$ such that $\cos x \neq 0$ .

Find the derivative of arccos (1/x)

by RoRi

November 29, 2015

Find the derivative of the function

$f(x) = \arccos \frac{1}{x}.$

By definition of arcsecant we know

$f(x) = \arccos \frac{1}{x} = \operatorname{arcsec} x.$

We computed the derivative of $\arcsec x$ in this exercise (Section 6.22, Exercise #4) so,

$f'(x) = \frac{1}{|x| \sqrt{x^2 - 1}}$

valid for $|x| > 1$ .

Find the derivative of arccos ((1-x)/2^1/2)

by RoRi

November 29, 2015

Find the derivative of the function

$f(x) = \arccos \frac{1-x}{\sqrt{2}}.$

Using the formula for the derivative of $\arccos x$ and the chain rule we have

$\begin{align*} f'(x) &= \left( \frac{-1}{\sqrt{2}} \right) \left( \frac{-1}{\sqrt{1- \left( \frac{1-x}{\sqrt{2}} \right)^2}}\right) \\ &= \frac{1}{\sqrt{2 \left( 1 - \frac{1}{2} + x - \frac{x^2}{2} \right)}} \\ &= \frac{1}{\sqrt{1+2x-x^2}}. \end{align*}$

Since the domain of $\arccos x$ is $|x| < 1$ , this formula is valid for $|x-1| < \sqrt{2}$ .

Find the derivative of arcsin (x/2)

by RoRi

November 29, 2015

Find the derivative of the function

$f(x) = \arcsin \frac{x}{2}.$

Using the chain rule and the derivative of arcsine we have

$f'(x) = \left( \frac{1}{2} \right) \left( \frac{1}{\sqrt{1- \left(\frac{x}{2}\right)^2}} \right) = \frac{1}{\sqrt{4-x^2}}.$

This is valid for $|x| < 2$ (since the domain of $\arcsin x$ is $|x| < 1$ ).

Prove that arccot x – arctan (1/x) is not constant but has zero derivative

by RoRi

November 29, 2015

Prove that for $x \neq 0$ we have
$D \left( \operatorname{arccot} x - \arctan \frac{1}{x} \right) = 0.$
Prove that there is no constant $C \in \mathbb{R}$ such that for all $x \neq 0$ we have
$\operatorname{arccot} x - \arctan \frac{1}{x} = C.$

Why is this not a violation of the zero-derivative theorem (Theorem 5.2 in Apostol)?

Proof. We can use the formulas for the derivatives of $\operatorname{arccot} x$ and $\arctan x$ (and the chain rule) to compute,
$\begin{align*} D \left( \operatorname{arccot} x - \arctan \frac{1}{x} \right) &= -\frac{1}{1+x^2} - \left( \frac{1}{1+\left( \frac{1}{x} \right)^2} \right) \left(\frac{-1}{x^2}\right) \\[9pt] &= \frac{-1}{1+x^2} + \frac{1}{1+x^2} \\[9pt] &= 0. \qquad \blacksquare \end{align*}$
Proof. First, let $x = -1$ . Then,
$\arctan x = \arctan \frac{1}{x} = \arctan(-1) = -\frac{\pi}{4}.$

Then, since $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$ , we have

$\operatorname{arccot} x - \arctan \frac{1}{x} = \frac{\pi}{2} - \arctan x - \arctan \frac{1}{x} = \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{4} = \pi.$

Next, let $x = 1$ . Then, $\arctan x = \arctan \frac{1}{x} = \arctan 1 = \frac{\pi}{4}.$ Again, using that $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$ , we have

$\operatorname{arccot} x - \arctan \frac{1}{x} = \frac{\pi}{2} - \frac{\pi}{4} - \frac{\pi}{4} = 0.$

Hence, there is no constant $C$ such that $\operatorname{arccot} x - \arctan \frac{1}{x} = C$ for all $x \neq 0. \qquad \blacksquare$

This is not a violation of the zero-derivative theorem since the function $\operatorname{arccot} x - \arctan \frac{1}{x}$ is constant on every open interval on which it is defined. Since it isn’t defined at $x = 0$ , any open subinterval must be a subinterval of only positive or only negative reals. The function is constant on any of these subintervals.

Establish the formula for the derivative of arccsc x

by RoRi

November 29, 2015

Establish the following formula for the derivative of $\arcsec x$ is correct,

$D(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}, \qquad \text{for all } |x| >1.$

For $x > 1$ let

$y = \operatorname{arccsc} x \quad \implies \quad x = \csc y.$

Then we know

$\frac{dx}{dy} = -\csc y \cot y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\csc y \cot y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\csc y \cot y} \\ && \implies && D(\operatorname{arccsc} x) &= \frac{-1}{\csc y \cot y} \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \left( \frac{-1}{\csc y \cot y} \right)^2 \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \frac{1}{\csc^2 y \cot^2 y} \\ && \implies && (D(\operatorname{arccsc} x))^2 &= \frac{1}{\csc^2 y (\csc^2 y - 1)}. \end{align*}$

Where we use the trig identity $\cot^2 y = \csc^2 y - 1$ in the final line. Then, since $x = \csc y$ we have,

$\begin{align*} &&(D(\operatorname{arccsc} x))^2 &= \frac{1}{x^2 (x^2 - 1)} \\ \implies && D (\arcsec x) &= \frac{-1}{|x|\sqrt{x^2-1}}. \qquad \blacksquare \end{align*}$

Establish the formula for the derivative of arcsec x

by RoRi

November 29, 2015

Establish the following formula for the derivative of $\arcsec x$ is correct,

$D(\operatorname{arcsec} x) = \frac{1}{|x|\sqrt{x^2-1}}, \qquad \text{for all } |x| >1.$

For $x > 1$ let

$y = \operatorname{arcsec} x \quad \implies \quad x = \sec y.$

Then we know

$\frac{dx}{dy} = \sec y \tan y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && \sec y \tan y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{1}{\sec y \tan y} \\ && \implies && D(\operatorname{arcsec} x) &= \frac{1}{\sec y \tan y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \left( \frac{1}{\sec y \tan y} \right)^2 \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y \tan^2 y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y (\sec^2 y - 1)}. \end{align*}$

Where we use the trig identity $\tan^2 y = \sec^2 y - 1$ in the final line. Then, since $x = \sec y$ we have,

$\begin{align*} &&(D(\operatorname{arcsec} x))^2 &= \frac{1}{x^2 (x^2 - 1)} \\ \implies && D (\operatorname{arcsec} x) &= \frac{1}{|x|\sqrt{x^2-1}}. \qquad \blacksquare \end{align*}$

Establish the formula for the derivative of arccot x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arctan x$ is correct,

$D(\operatorname{arccot} x) = \frac{-1}{1+x^2}, \qquad \text{for all } x \in \mathbb{R}.$

For $x \in \mathbb{R}$ let

$y = \operatorname{arccot} x \quad \implies \quad x = \cot y.$

Then we know

$\frac{dx}{dy} = -\csc^2 y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\csc^2 y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\csc^2 y} \\ && \implies && D(\operatorname{arccot} x) &= \frac{-1}{\csc^2 y}. \end{align*}$

Using a trig identity for tangent and cosecant we have

$\csc^2 y = 1 + \cot^2 y = 1 + x^2$

since $x = \cot y$ . Therefore we conclude,

$D(\operatorname{arccot} x) = \frac{-1}{1+x^2}.$

Establish the formula for the derivative of arctan x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arctan x$ is correct,

$D(\arctan x) = \frac{1}{1+x^2}, \qquad \text{for all } x \in \mathbb{R}.$

For $x \in \mathbb{R}$ let

$y = \arctan x \quad \implies \quad x = \tan y.$

Then we know

$\frac{dx}{dy} = \sec^2 y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && \sec^2 y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{1}{\sec^2 y} \\ && \implies && D(\arctan x) &= \frac{1}{\sec^2 y}. \end{align*}$

Using a trig identity for tangent and secant we have

$\sec^2 y = 1 + \tan^2 y = 1 + x^2$

since $x = \tan y$ . Therefore we conclude,

$D(\arctan x) = \frac{1}{1+x^2}.$

Establish the formula for the derivative of arccos x

by RoRi

November 23, 2015

Establish the following formula for the derivative of $\arccos x$ is correct,

$D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}, \qquad \text{if } -1<x<1.$

For $-1<x<1$ let

$y = \arccos x \quad \implies \quad x = \cos y.$

Then we know

$\frac{dx}{dy} = -\sin y.$

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

$\begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\sin y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\sin y} \\ && \implies && D(\arccos x) &= \frac{-1}{\sin y}. \end{align*}$

Using the pythagorean identity for sine and cosine we have

$\begin{align*} \sin^2 y + \cos^2 y =1 && \implies && \sin y &= \pm \sqrt{1-\cos^2 y} \\ && \implies && \sin y &= \pm \sqrt{1-x^2} \end{align*}$

since $x = \cos y$ . Furthermore, since $\sin y \geq 0$ on the range of $\arccos x$ (i.e., $\sin y \ge 0$ for $y \in [0, \pi]$ ) we must take the positive square root. Therefore we conclude,

$D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}.$