The derivative of a square root function

Algebra and calculus, grade 10 (4 units) Lesson 17 of 20Learning

The derivative of a square root function

The domain of

\displaystyle \sqrt{f(x)}

The function $\displaystyle {y}={\sqrt{f(x)}}$ is defined only where $\displaystyle f(x)\geq0$ , because a negative number has no square root.

$\displaystyle {f(x)}={\sqrt{12-3x}}$ : solving $\displaystyle 12-3x\geq0$ gives $\displaystyle x\leq4$ .
$\displaystyle {f(x)}={\sqrt{x^2-4x+3}}$ : the roots of $\displaystyle {x^2-4x+3}={0}$ are 1 and 3, and the parabola opens upwards – it is above the $\displaystyle x$ -axis on both sides of the roots. The domain: $\displaystyle x\leq1$ or $\displaystyle x\geq3$ .
At an end of the domain the value of the function is 0: $\displaystyle {\sqrt{0}}={0}$ .

The derivative of

\displaystyle \sqrt{x}

The derivative of $\displaystyle \sqrt{x}$ is $\displaystyle \frac{1}{2\sqrt{x}}$ .

\displaystyle (\sqrt{x})'

\displaystyle =

\displaystyle \frac{1}{2\sqrt{x}}

\displaystyle x>0

The calculation below finds it from the slope of the chord from $\displaystyle x$ to $\displaystyle x+h$ . We multiply the numerator and the denominator by $\displaystyle \sqrt{x+h}+\sqrt{x}$ , so that the numerator becomes a difference of squares:

\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\frac{1}{\sqrt{x+h}+\sqrt{x}}

As $\displaystyle h$ gets closer to 0, $\displaystyle \sqrt{x+h}$ gets closer to $\displaystyle \sqrt{x}$ , and the slope gets closer to $\displaystyle \frac{1}{2\sqrt{x}}$ .
At $\displaystyle {x}={0}$ the function has no derivative: the tangent at $\displaystyle (0,0)$ is the $\displaystyle y$ -axis, a line upright to the $\displaystyle x$ -axis, which has no slope.

The derivative of

\displaystyle \sqrt{f(x)}

$\displaystyle \sqrt{f(x)}$ is a composite function: the outer function is the root, and the inner one is $\displaystyle f(x)$ . So we differentiate the root and multiply by the derivative of the inner one:

\displaystyle (\sqrt{f(x)})'

\displaystyle =

\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}

\displaystyle f(x)>0

The examples below differentiate two such functions:

\left(\sqrt{x^2+9}\right)'=\frac{2x}{2\sqrt{x^2+9}}=\frac{x}{\sqrt{x^2+9}}

\left(\sqrt{12-3x}\right)'=\frac{-3}{2\sqrt{12-3x}}

Sums and products of a root and a polynomial

In a sum each term is differentiated on its own; in a product we use the rule for the derivative of a product.

\left(\sqrt{2x+1}+x^2\right)'=\frac{2}{2\sqrt{2x+1}}+2x=\frac{1}{\sqrt{2x+1}}+2x

The example below differentiates $\displaystyle x\sqrt{x+5}$ as a product, and brings the result to a common denominator:

\left(x\sqrt{x+5}\right)'=1\cdot\sqrt{x+5}+x\cdot\frac{1}{2\sqrt{x+5}}=\frac{2(x+5)+x}{2\sqrt{x+5}}=\frac{3x+10}{2\sqrt{x+5}}

$\displaystyle \sqrt{x+5}$ is multiplied by $\displaystyle 2\sqrt{x+5}$ to bring it over the denominator, which gives $\displaystyle 2(x+5)$ .
A common denominator helps later: the equation $\displaystyle {f'(x)}={0}$ is solved by the numerator alone.

The tangent to the graph of a root function

The tangent to the graph of a square root function is found just as for a polynomial: the point, the slope $\displaystyle {m}={f'(x_1)}$ , and then $\displaystyle {y-y_1}={m(x-x_1)}$ .

The functions are $\displaystyle {f(x)}={\sqrt{12-3x}}$ and $\displaystyle {g(x)}={-f(x)}$ .
The example below finds the tangent to each graph at the point where $\displaystyle {x}={1}$ , and the point where the tangents meet.

$\displaystyle {f(1)}={\sqrt{9}}={3}$ , and the slope is $\displaystyle {f'(1)}={\frac{-3}{2\cdot3}}={-\frac{1}{2}}$ .
The tangent to the graph of $\displaystyle f$ : $\displaystyle {y-3}={-\frac{1}{2}(x-1)}$ , that is $\displaystyle {y}={-\frac{1}{2}x+3\frac{1}{2}}$ .
The graph of $\displaystyle g$ is the reflection of the graph of $\displaystyle f$ in the $\displaystyle x$ -axis, so the tangent is reflected too: $\displaystyle {g(1)}={-3}$ , $\displaystyle {g'(1)}={\frac{1}{2}}$ , and the tangent is $\displaystyle {y}={\frac{1}{2}x-3\frac{1}{2}}$ .

\displaystyle f(x)=\sqrt{12-3x}

\displaystyle g(x)=-\sqrt{12-3x}

\displaystyle y=-\frac{1}{2}x+\frac{7}{2}

\displaystyle y=\frac{1}{2}x-\frac{7}{2}

\displaystyle A=(7,0)

The tangents meet where $\displaystyle {-\frac{1}{2}x+3\frac{1}{2}}={\frac{1}{2}x-3\frac{1}{2}}$ , that is at $\displaystyle {x}={7}$ .
The meeting point $\displaystyle (7,0)$ is on the $\displaystyle x$ -axis, the axis of symmetry of the two graphs.

Exercise 1

Find the domain of the function.

$\displaystyle {f(x)}={\sqrt{2x-6}}$

$\displaystyle {f(x)}={\sqrt{16-x^2}}$

Exercise 2

Find the derivative.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The function is $\displaystyle {f(x)}={\sqrt{2x+1}}$ .
Find $\displaystyle f'(4)$ .

$\displaystyle f(x)=\sqrt{2x+1}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

The function is $\displaystyle {f(x)}={\sqrt{x+3}}$ .
The point where $\displaystyle {x}={1}$ is marked on the graph.
Write on the graph board the equation of the tangent to the graph at this point.

\displaystyle f(x)=\sqrt{x+3}

\displaystyle (1,2)

Exercise 5

The function is $\displaystyle {f(x)}={\sqrt{2x-1}}$ .
At a certain point of the graph the slope of the tangent is $\displaystyle \frac{1}{3}$ .

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\sqrt{2x-1}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Write an equation and solve it: what is the $\displaystyle x$ -coordinate of the point?

What are the coordinates of the point?

Exercise 6

The functions are $\displaystyle {f(x)}={\sqrt{8-2x}}$ and $\displaystyle {g(x)}={-f(x)}$ .

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\sqrt{8-2x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Write on the graph board the tangent to the graph of $\displaystyle f$ and the tangent to the graph of $\displaystyle g$ , each at the point where $\displaystyle {x}={2}$ .

\displaystyle f(x)=\sqrt{8-2x}

\displaystyle g(x)=-\sqrt{8-2x}

Write an equation and solve it: what is the $\displaystyle x$ -coordinate of the point where the tangents meet?

Find the area of the triangle formed by the two tangents and the line $\displaystyle {x}={2}$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: algebra and calculus, grade 10 (4 units)

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Algebraic technique

The square of a sum
$\displaystyle (a+b)^2$ $\displaystyle =$ $\displaystyle a^2+2ab+b^2$
The square of a difference
$\displaystyle (a-b)^2$ $\displaystyle =$ $\displaystyle a^2-2ab+b^2$
The difference of squares
$\displaystyle (a+b)(a-b)$ $\displaystyle =$ $\displaystyle a^2-b^2$
Factoring a trinomial – $\displaystyle x_1$ , $\displaystyle x_2$ the roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle ax^2+bx+c$ $\displaystyle =$ $\displaystyle a(x-x_1)(x-x_2)$
The roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The product of roots
$\displaystyle \sqrt{a}\cdot\sqrt{b}$ $\displaystyle =$ $\displaystyle \sqrt{ab}$ $\displaystyle a\geq0,\ b\geq0$
The quotient of roots
$\displaystyle \frac{\sqrt{a}}{\sqrt{b}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{a}{b}}$ $\displaystyle a\geq0,\ b>0$

The properties of a function

An even function – its graph symmetric about the $\displaystyle y$ -axis
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle f(x)$
An odd function – its graph symmetric about the origin
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle -f(x)$
The $\displaystyle x$ -coordinate of the vertex of $\displaystyle {y}={ax^2+bx+c}$
$\displaystyle x=\frac{-b}{2a}$
The absolute value
$\displaystyle |a|=\begin{cases}a & a\geq0\\-a & a<0\end{cases}$

Transforming a graph

A shift of $\displaystyle k$ units up ( $\displaystyle k<0$ – down)
$\displaystyle y=f(x)+k$
A shift of $\displaystyle p$ units to the right ( $\displaystyle p<0$ – to the left)
$\displaystyle y=f(x-p)$
A reflection in the $\displaystyle x$ -axis
$\displaystyle y=-f(x)$
A reflection in the $\displaystyle y$ -axis
$\displaystyle y=f(-x)$
A stretch ( $\displaystyle |a|>1$ ) or a shrink ( $\displaystyle |a|<1$ )
$\displaystyle y=a\cdot f(x)$

The derivative and the rules of differentiation

The average rate of change from $\displaystyle x_1$ to $\displaystyle x_2$ – the slope of the chord
$\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$
The derivative at a point – the slope of the tangent
$\displaystyle f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$
The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
The derivative of a constant
$\displaystyle (c)'$ $\displaystyle =$ $\displaystyle 0$
The derivative of a constant multiple
$\displaystyle (c\cdot f(x))'$ $\displaystyle =$ $\displaystyle c\cdot f'(x)$
The derivative of a sum and of a difference
$\displaystyle (f(x)\pm g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\pm g'(x)$
The derivative of a product
$\displaystyle (f(x)\cdot g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\cdot g(x)+f(x)\cdot g'(x)$
The derivative of a composite function
$\displaystyle (f(g(x)))'$ $\displaystyle =$ $\displaystyle f'(g(x))\cdot g'(x)$

The tangent to a graph

The slope of the tangent at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle m=f'(x_1)$
The equation of the tangent
$\displaystyle y-y_1=m(x-x_1)$

The derivative of a square root function

The derivative of $\displaystyle \sqrt{x}$
$\displaystyle (\sqrt{x})'$ $\displaystyle =$ $\displaystyle \frac{1}{2\sqrt{x}}$ $\displaystyle x>0$
The derivative of the root of a function
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Rising, falling and extremum points

The function rises
$\displaystyle f'(x)>0$
The function falls
$\displaystyle f'(x)<0$
A point that may be an extremum
$\displaystyle f'(x)=0$