Derivation rules arcsin

Let $\displaystyle f(x)= a -2 \arcsin(1-2x)$ with $a \in \mathbb{R}$ and let $f'(x)$ and $f''(x)$ be the first and the second derivative of a function $f$. Knowing that $\displaystyle \dfrac{f\left(\dfrac{1}{2}\right)}{f'\left(\dfrac{1}{2}\right)} + f''\left(\frac{1}{2}\right) = 2$, then the value of $a$ is:

# Functions of one variable

Training
13

Tangent and normal lines

The figure shows : - part of a function $f$ - a line $r$ tangent to the graph of $f$ at the point with abscissa iqual to $2$ The value of $f'(2)$, derivative of the function $f$ at the point of abscissa $2$, can be equal to:

# Functions of one variable

Reasoning
13

Derivation rules arcsin

Consider $f(x)= 2 \arcsin\left(\dfrac{x}{3}\right)$ with domain $D_f=[-3, 3]$. Let $f'(x)$ be the first derivative of function $f(x)$, then $f'\left(1\right)$ is:

# Functions of one variable

Training
13

Derivation rules arcsin

Let $f$ and $g$ be two differentiable functions in their domain, such that: $f(x)= \dfrac{\pi}{2} -2 \arcsin(1-2x)$ and $g(x)=x^2 f(x)$. Considering that $g'(x)$ is the first derivative of function $g(x)$, then $g'\left(\dfrac{1}{2}\right)$ is:

# Functions of one variable

Training
13

Solve an equation or inequality with arcsin

Consider $f(x)= 2 \arcsin\left(\dfrac{x}{3}\right)$ with domain $D_f=[-3, 3]$. The zeros of $f(x)$ are:

# Functions of one variable

Training
13

Solve an equation or inequality with arcsin

Consider $f(x)= \dfrac{\pi}{2} -2 \arcsin(1-2x)$ with domain $D_f=[0, 1]$. The solution of the equation $f(x)=f(0) + 2 \arcsin\left( \dfrac{\sqrt{3}}{2}\right)$ is:

# Functions of one variable

Training
13