## The First Derivatives

 2.2 The First Derivatives

• For the function $$y=ax^n$$, where $$a$$ is a constant and $$n$$ is an integer:
 \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}

 For $$y=ax^n$$, If $$n=1, \ \dfrac{dy}{dx}=a$$ If $$n=0, \ \dfrac{dy}{dx}=0$$

$$3$$ 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}

• Chain Rule:
 $$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$

 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}

• Product Rule:
 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 $$4$$.

• Quotient Rule:
 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}

## The First Derivatives

 2.2 The First Derivatives

• For the function $$y=ax^n$$, where $$a$$ is a constant and $$n$$ is an integer:
 \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}

 For $$y=ax^n$$, If $$n=1, \ \dfrac{dy}{dx}=a$$ If $$n=0, \ \dfrac{dy}{dx}=0$$

$$3$$ 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}

• Chain Rule:
 $$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$

 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}

• Product Rule:
 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 $$4$$.

• Quotient Rule:
 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}