Algebraic Functions: First Derivative

An equation that has only a single answer for the "y" in every "x", is a function. Function has only one output as a result to every input in the equation, y = x, or, y = f(x)= x. Some symbols simplified written as f(x), g(x), h(x), etc. The given f(x) = x+ 2, when f(1) is asked for the function, it means that, we wanna find the function when "x" is at 1.

In differential calculus, we find their derivatives. We use "y' " for f '(x), "y' read as 'y prime' or y first derivative", or sometimes , we found them with their first derivative as;

${y}'=D_{x}f({\color{Red} x})=\frac{d}{dx}f({\color{Red} x})=\frac{df}{dx}x=f{}'{\color{Red} x}=f^{1}{\color{Red} x}; {\color{Red} y}=f({\color{Red} x})$

We knew that in differential calculus, we have the following rules of derivatives like;

$i). \frac{d{\color{Red} c}}{dx}=0;{\color{Red} c} = constant$

$ii). \frac{d}{dx}(\frac{{\color{Red} u}}{{\color{Purple} v}})=\frac{d{\color{Red} u}}{dx}+\frac{d{\color{Purple} v}}{dx}$

$iii). \frac{d}{dx}({\color{Red} u}{\color{Blue} v})={\color{Red} u}\frac{dv}{dx}+{\color{Blue} v}\frac{du}{dx}$ ; (u and v are function of x)

$iv).\frac{d}{dx}(\frac{{\color{Red} u}}{{\color{Purple} v}})=\frac{{\color{Purple} v}\frac{du}{dx}-{\color{Red} u}\frac{dv}{dx}}{{\color{Purple} v}^{2}}$

$v). \frac{d}{dx}(x^{{\color{Red} n}})={{\color{Red} n}}x^{{{\color{Red} n}}-1}$

Now, see our illustrative examples.

Illustrative Examples

$1). y=3x^{2}$

   ${y}'=2(3)x^{2-1}$

   ${y}'=6x$ , ans

$2.) y=4x^{3}-2x+3$

   ${y}'=3(4)x^{3-1}-2x^{1-1} +3$

   ${y}'=12x^{2}-2x^{0}+0$

   ${y}'=12x^{2}-2$ , ans

$3.)y=\frac{2}{x}$

   $y=2x^{-1}$

${y}'=-1(2)x^{-1-1}$

${y}'=-2x^{-2}$

${y}'=\frac{-2}{x^{2}}$ , ans

$4.)y=\frac{2}{x}+\frac{1}{x^{2}}$

   $y=2x^{-1} +x^{-2}$

   ${y}'=-1(2)x^{-1-1}+(-2)x^{-2-1}$

   ${y}'=-2x^{-2}-2x^{-3}$

   ${y}'=-\frac{2}{x^{2}}-\frac{2}{x^{3}}$ , ans

Sample Problems with solution:

Differentiate the following Algebraic Functions.

$1.)y=3x^{2}-2x+3$

${y}'=(2)3x^{2-1}-(1)2x^{1-1}+0$

   ${y}'=6x^{1}-2x^{0}$

   ${y}'=6x-2$ , ans

$2.)y=2x^{4}-3x^{3}+2$

${y}'=(4)2x^{4-1}-3(3)x^{3-1}+0$

${y}'=8x^{3}-9x^{2}$ , ans

$3.)y= x^{3}+5x^{2}+2x+1$

${y}'=(3)x^{3-1}+(2)5x^{2-1}+2x^{1-1}+0$

${y}'=3x^{2}+10x+2(1)$

${y}'=3x^{2}+10x+2$ , ans

$4.)y=7-4x^{2}-3x^{3}+x^{4}$

${y}'=0-(2)4x^{2-1}-(3)3x^{3-1}+4x^{4-1}$

${y}'=-8x-9x^{2}+4x^{3}$ , ans

$5.) y= \frac{2}{x} -\frac{3}{x^{2}}$

$y=2x^{-1}-3x^{-2}$

${y}'=(-1)2x^{-1-1}-(-2)3x^{-2-1}$

   ${y}'=-2x^{-2}+6x^{-3}$

   ${y}'=-\frac{2}{x^{2}} +\frac{6}{x^{3}}$ , ans

Pop Quiz:

Differentiate the following, with partial solution:

$1.) y= 3x^{2}+1$

${y}'=(2)3x^{2-1}+0$ , partial solution

${y}'=$ click here to write your final answer

$2.) y=3x^{3}-2x^{2}$

${y}'=(3)3x^{2-1}-(2)2x^{2-1}$ , partial solution

   ${y}'= 9x^{2}-$ click here to write your final answer

$3. y=\frac{3}{x^{3}}-2\frac{1}{x^{2}}$

   ${y}'=3x^{-3}-2x^{-2}$ , partial solution

   ${y}'=$ click here to write your final answer

TAKE A SHORT QUIZ:

Differentiate the following Algebraic Functions. (YouTube solution soon)

$1.) y=5x^{3}+2x$

$2.) y=3x^{2}-4x+3$

$3.) y=6+3x^{2}-2x^{4}$

$4.) y= 2\frac{1}{x^{2}}+\frac{2}{x}$

$5.) y= \frac{3}{x^{2}}-\frac{5}{x}+1$

More exercises on derivative:

If you see typographical or any error(s), please feel free "Post a comment" below.

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DMG

Algebraic Functions: First Derivative

Posted by : Allan_Dell on Monday, April 6, 2020 | 12:34 AM

+ comments + 1 comments

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DMG

Algebraic Functions: First Derivative

Posted by : Allan_Dell on Monday, April 6, 2020 | 12:34 AM

+ comments + 1 comments

Popular posts

Posted by : Allan_Dell on Monday, April 6, 2020 | 12:34 AM