Derivatives of functions with roots, and root equations

Calculus, grade 11 (5 units) Lesson 14 of 26Learning

Derivatives of functions with roots, and root equations

A product and a quotient with a root

A function with a root is differentiated by the rules you know, with the derivative of the root of a function:

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

\displaystyle =

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

In the two examples below, the last step brings the derivative over one denominator, where its sign is easy to read:

\left(x\sqrt{4-x^2}\right)'=\sqrt{4-x^2}+x\cdot\frac{-2x}{2\sqrt{4-x^2}}=\frac{(4-x^2)-x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}

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

The first derivative is defined where $\displaystyle -2<x<2$ : at the ends $\displaystyle \pm2$ of the domain the root in its denominator is 0.
The second is defined where $\displaystyle x>0$ : the function is defined at 0, its derivative is not.

A vertical tangent at the end of the domain

At an end of the domain where a root is 0, the derivative usually has no value, and the slopes near it grow without end. For $\displaystyle {f(x)}={\sqrt{x}}$ :

$\displaystyle {f'(x)}={\frac{1}{2\sqrt{x}}}$ : at $\displaystyle {x}={0.01}$ the slope is 5, at $\displaystyle {x}={0.0001}$ it is 50;
the tangents get steeper and steeper as $\displaystyle x$ approaches 0, and the tangent at $\displaystyle (0,0)$ is the $\displaystyle y$ -axis – a vertical tangent;
the graph of $\displaystyle f'$ has the vertical asymptote $\displaystyle {x}={0}$ .

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

\displaystyle g(x)=\frac{1}{2\sqrt{x}}

Equations with a root

An equation with a root is solved by squaring both sides, and squaring may add solutions: $\displaystyle {a}={b}$ and $\displaystyle {a}={-b}$ both give $\displaystyle {a^2}={b^2}$ . So every solution of the squared equation is checked in the original one. The example below solves $\displaystyle {\sqrt{x+3}}={x+1}$ :

squaring: $\displaystyle {x+3}={x^2+2x+1}$ , that is $\displaystyle {x^2+x-2}={0}$ , so $\displaystyle {x}={1}$ or $\displaystyle {x}={-2}$ ;
$\displaystyle {x}={1}$ : $\displaystyle {\sqrt{4}}={2}$ and $\displaystyle {1+1}={2}$ – a solution;
$\displaystyle {x}={-2}$ : $\displaystyle {\sqrt{1}}={1}$ but $\displaystyle {-2+1}={-1}$ – not a solution: it solves $\displaystyle {\sqrt{x+3}}={-(x+1)}$ .

A shorter check: the root is never negative, so a solution must have $\displaystyle x+1\geq0$ .

Taking

\displaystyle x

into the root

Noam investigated $\displaystyle {f(x)}={x\sqrt{4-x^2}}$ . To save the product rule, he wrote $\displaystyle {f(x)}={\sqrt{x^2(4-x^2)}}={\sqrt{4x^2-x^4}}$ , and his sketch had no negative values at all. The step is right only where $\displaystyle x\geq0$ :

$\displaystyle {\sqrt{x^2(4-x^2)}}={|x|\sqrt{4-x^2}}$ , since $\displaystyle {\sqrt{x^2}}={|x|}$ ;
for $\displaystyle x<0$ , $\displaystyle x\sqrt{4-x^2}$ is negative, while $\displaystyle \sqrt{4x^2-x^4}$ is not;
so $\displaystyle \sqrt{4x^2-x^4}$ is a different function: its graph is that of $\displaystyle f$ with the part left of the $\displaystyle y$ -axis flipped above the $\displaystyle x$ -axis.

Exercise 1

Differentiate, and bring each derivative over one denominator.

/ – 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 2

The equation is $\displaystyle {\sqrt{2x+7}}={x+2}$ .

Square both sides, and solve the equation you get.

Check each solution in the original equation, and write here only the solutions of the original equation.

Exercise 3

The function is $\displaystyle {f(x)}={x\sqrt{x+3}}$ .
The point $\displaystyle (1,2)$ is on its graph.

Work out the slope of the tangent at the point $\displaystyle (1,2)$ .

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

/ – 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 the equation of the tangent on the board.

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

\displaystyle (1,2)

Exercise 4

The function is $\displaystyle {f(x)}={\sqrt{9-x^2}}$ .

Find $\displaystyle f'(x)$ .

$\displaystyle f(x)=\sqrt{9-x^{2}}$

/ – 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

What happens to $\displaystyle f'(x)$ as $\displaystyle x$ approaches 3 from the left?
What is the tangent to the graph at the point $\displaystyle (3,0)$ ?

Exercise 5

Dana wrote: $\displaystyle {\sqrt{x^2-6x+9}}={x-3}$ .
For which $\displaystyle x$ is it true? Explain by $\displaystyle {\sqrt{m^2}}={|m|}$ .

Formulas: calculus, grade 11 (5 units)

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Rules of differentiation and the tangent line

The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
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 quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]'$ $\displaystyle =$ $\displaystyle \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$
The derivative of a composite function – the outer derivative times the inner derivative
$\displaystyle [f(u(x))]'$ $\displaystyle =$ $\displaystyle f'(u)\cdot u'(x)$
The derivative of $\displaystyle \frac{1}{x}$
$\displaystyle \left(\frac{1}{x}\right)'$ $\displaystyle =$ $\displaystyle -\frac{1}{x^2}$
The tangent line at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle y-y_1=f'(x_1)(x-x_1)$

The second derivative

The second derivative – the derivative of the derivative
$\displaystyle f''(x)$ $\displaystyle =$ $\displaystyle \left(f'(x)\right)'$
The function is concave up ( $\displaystyle \cup$ )
$\displaystyle f''(x)>0$
The function is concave down ( $\displaystyle \cap$ )
$\displaystyle f''(x)<0$
A possible inflection point – check that the sign of $\displaystyle f''$ changes there
$\displaystyle f''(x_0)=0$
A minimum point
$\displaystyle f'(x_0)=0,\ f''(x_0)>0$
A maximum point
$\displaystyle f'(x_0)=0,\ f''(x_0)<0$
Acceleration – the derivative of the speed
$\displaystyle a(t)=v'(t)=s''(t)$

Asymptotes

A vertical asymptote – $\displaystyle f(x)$ grows or falls without end as $\displaystyle x$ approaches $\displaystyle a$
$\displaystyle x=a$
A horizontal asymptote – $\displaystyle f(x)$ approaches $\displaystyle b$ as $\displaystyle x\to\infty$ or as $\displaystyle x\to-\infty$
$\displaystyle y=b$

Roots

The root of a square
$\displaystyle \sqrt{x^2}$ $\displaystyle =$ $\displaystyle |x|$
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
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Radians and periods

Half a turn in radians
$\displaystyle \pi$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The length of an arc of $\displaystyle \alpha$ radians in a circle of radius $\displaystyle R$
$\displaystyle l$ $\displaystyle =$ $\displaystyle \alpha R$
The period of $\displaystyle \sin x$ and of $\displaystyle \cos x$ is $\displaystyle 2\pi$
$\displaystyle \sin(x+2\pi)$ $\displaystyle =$ $\displaystyle \sin x$
The period of $\displaystyle \tan x$ is $\displaystyle \pi$
$\displaystyle \tan(x+\pi)$ $\displaystyle =$ $\displaystyle \tan x$
The limit behind the derivative of $\displaystyle \sin x$ ( $\displaystyle x$ in radians)
$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$

Trigonometric identities

The Pythagorean identity
$\displaystyle \sin^2\alpha+\cos^2\alpha$ $\displaystyle =$ $\displaystyle 1$
Tangent
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
Sine – an odd function
$\displaystyle \sin(-\alpha)$ $\displaystyle =$ $\displaystyle -\sin\alpha$
Cosine – an even function
$\displaystyle \cos(-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
The sine of the supplement to $\displaystyle \pi$
$\displaystyle \sin(\pi-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The cosine of the complement to $\displaystyle \frac{\pi}{2}$
$\displaystyle \cos\left(\frac{\pi}{2}-\alpha\right)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of a double angle
$\displaystyle \sin2\alpha$ $\displaystyle =$ $\displaystyle 2\sin\alpha\cdot\cos\alpha$
The cosine of a double angle
$\displaystyle \cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$
The difference of two sines
$\displaystyle \sin\alpha-\sin\beta$ $\displaystyle =$ $\displaystyle 2\cos\frac{\alpha+\beta}{2}\cdot\sin\frac{\alpha-\beta}{2}$
The sine of a sum and of a difference
$\displaystyle \sin(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta$

Derivatives of the trigonometric functions

The derivative of sine ( $\displaystyle x$ in radians)
$\displaystyle (\sin x)'$ $\displaystyle =$ $\displaystyle \cos x$
The derivative of cosine
$\displaystyle (\cos x)'$ $\displaystyle =$ $\displaystyle -\sin x$
The derivative of tangent
$\displaystyle (\tan x)'$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^2x}$