The First Derivatives

2.2

The First Derivatives


	\( \text{If }y=ax^n, \text{then} \\\\ \begin{aligned} \dfrac{dy}{dx} = anx^{n-1} \text{ or } \\\\\dfrac{d}{dx}(ax^n) = anx^{n-1} \end{aligned}\)

\(3\) Notations Used to Indicate the First Derivative of a Function \(y=ax^n\)

1.	If \(y=3x^2, \text{ then } \dfrac{dy}{dx} = 6x\)

	\(\dfrac{dy}{dx}\) is read as differentiating \(y \text{ with respect to } x\)

2.	If \(f(x) =3x^2, \text{ then } f'(x)=6x\)

	\(f'(x)\) is known as the gradient function for the curve \(y=f(x)\) at any point on the curve

3.	\(\dfrac{dy}{dx}(3x^2) = 6x\)

	If differentiating \(3x^2 \text{ with respect to } x, \text{ the result is } 6x\)

Differentiation

The process of determining the gradient function \(f'(x)\) from

a function \(y=f(x)\)

The gradient function is also known as the first derivative of the function or the derivatived function or differentiating coefficient of \(y\) with respect to \(x\).

Example

Differentiate each of the following with respect to \(x\) :

(a)	\(y = \dfrac{1}{5}\sqrt x\)

(b)	If \(f(x) =\dfrac{3}{4}x^4, \text{ find } f'(-1) \text{ and } f'\begin{pmatrix} 1 \\ 3 \end{pmatrix}\)


Solution:

(a)	\(\begin{aligned} y &= \dfrac{1}{5}\sqrt x\\\\ &= \dfrac{1}{5}x^{\frac{1}{2}}\\\\ \dfrac{dy}{dx} &= \dfrac{1}{5} \begin{pmatrix} {\dfrac{1}{2}x^{\frac{1}{2}-1}} \end{pmatrix} \\\\ &=\dfrac{1}{10}x^{-\frac{1}{2}}\\\\ &= \dfrac{1}{10 \sqrt x} \end{aligned}\)

(b)	\(\begin{aligned} f(x) &=\dfrac{3}{4}x^4\\\\ f'(x) &= \dfrac{3}{4} (4x^{4-1})\\\\ &= 3x^3\\\\ f'(-1) &= 3(-1)^3\\ &=-3\\\\ f'\begin{pmatrix} 1 \\ 3 \end{pmatrix}&=3\begin{pmatrix} 1 \\ 3 \end{pmatrix}^3\\\\ &=\dfrac{1}{9} \end{aligned}\)

The derivative of a function which contains terms algebraically added or substracted can be done by differentiating each term separately
If \(f(x)\) and \(g(x)\) are functions, then:


	\( \dfrac{d}{dx}[f(x) \pm g(x)]\ = \ \dfrac{d}{dx}[f(x)] \pm \dfrac{d}{dx}[ g(x)]\ \)

	Example

	Differentiate each of the following with respect to \(x\) :
	\(\dfrac{(2x+1)(x-1)}{x}\)

	Solution:

	\(\begin{aligned} \text{Let } y &= \dfrac{(2x+1)(x-1)}{x}\\\\ &= \dfrac{2x^2-x-1}{x}\\\\ &=2x-1-x^{-1}\\\\\\ \dfrac{dy}{dx} &= \dfrac{d}{dx}(2x) -\dfrac{d}{dx}(1)-\dfrac{d}{dx}(x^{-1})\\\\ &=2x^{1-1} - 0x^{0-1} - (-1x^{-1-1})\\\\ &=2+x^{-2}\\\\ &= 2+ \dfrac{1}{x^2} \end{aligned}\)


	\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)


	If \(y=g(u)\) and \(u=h(x)\), then differentiating \(y\) with respect to \(x\) will give

	\(\begin{aligned} f'(x)&=g'(u) \times h'(x), \text{ that is}\\\\ \dfrac{dy}{dx} &= \dfrac{dy}{du} \times \dfrac{du}{dx} \end{aligned}\)

	Example

	Differentiate each of the following with respect to \(x\) :
	\(y=(3x^2-4x)^7\)

	Solution:

	\( \text{Let } u = (3x^2-4x) \text{ and } y=u^7\\\\\\ \text{Then, }\dfrac{du}{dx} = 6x-4 \text{ and } \dfrac{dy}{du} = 7u^6\\\\\\ \)
	With chain rule,

	\(\begin{aligned} \dfrac{dy}{dx} &= \dfrac{dy}{du} \times \dfrac{du}{dx}\\\\ &=7u^6(6x-4)\\\\ &=7(3x^2-4x)^6(6x-4)\\\\ &= (42x-28)(3x^2-4x)^6\\\\ &=14(3x-2)(3x^2-4x)^6 \end{aligned}\)


	If \(u\) and \(v\) are functions of \(x\), then

	\(\dfrac{d}{dx}(uv) = u \dfrac{dv}{dx} + v \dfrac{du}{dx}\)

Remark


	\(\dfrac{d}{dx}(uv) \neq \dfrac{du}{dx} \times \dfrac{dv}{dx}\)

Example

Given \(y=x \sqrt{x+3}\), find

(a)	the expression for \(\dfrac{dy}{dx}\)

(b)	the gradient of the tangent at \(x=6\)


Solution:

(a)	\(\text{Let }u=x \text{ and } v= \sqrt{x+3}\)

	Then using the product rule,

	\(\begin{aligned} \dfrac{dy}{dx} &= u \dfrac{dv}{dx} + v \dfrac{du}{dx}\\\\ &= x \dfrac{d}{dx} (\sqrt{x+3})+ \sqrt{x+3} \dfrac{d}{dx}x\\\\ &= x \begin{pmatrix} \dfrac{1}{2 \sqrt{x+3}}\end{pmatrix} + \sqrt{x+3} \\\\ &= \dfrac{x+2(x+3)}{2\sqrt{x+3} }\\\\ &= \dfrac{3(x+2)}{2\sqrt{x+3} } \end{aligned}\)

(b)	When \(x=6\),

	\(\begin{aligned} \dfrac{dy}{dx} &= \dfrac{3(6+2)}{2\sqrt{6+3} }\\\\ &=\dfrac{24}{6}\\\\ &= 4 \end{aligned}\)

	Hence, the gradient of the tangent at \(x=6\) is \(4\).


	If \(u\) and \(v\) are functions of \(x\) and \(v(x) \neq 0\), then

	\(\dfrac{d}{dx}\begin{pmatrix} \dfrac{u}{v} \end{pmatrix} = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}\)

Remark


	\(\dfrac{d}{dx}\begin{pmatrix} \dfrac{u}{v} \end{pmatrix} \neq \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} \)

	Example

	Given \(y=\dfrac{2x+1}{x^2-3},\text{ find } \dfrac{dy}{dx}\).

	Solution:

	\(\text{Let }u=2x+1 \text{ and } v= x^2-3\)

	Then, \(\dfrac{du}{dx} = 2 \text{ and } \dfrac{dv}{dx} = 2x\)

	\(\begin{aligned} \dfrac{dy}{dx} &= \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}\\\\ &= \dfrac{(x^2-3)(2)-(2x+1)(2x)}{(x^2-3)^2}\\\\ &= \dfrac{2x^2-6-(4x^2+2x)}{(x^2-3)^2}\\\\ &= \dfrac{-2x^2-2x-6}{(x^2-3)^2}\\\\ &= \dfrac{-2(x^2+x+3)}{(x^2-3)^2} \end{aligned}\)