Difference between revisions of "Simplifying Derivatives"

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<math>a'(x) =(3x^2)({(2x-5)}^4)  + (x^3)(8 {(2x-5)}^3)</math>
<math>a'(x) =(3x^2)({(2x-5)}^4)  + (x^3)(8 {(2x-5)}^3)</math>
factorize  
factorize  
<math>a'(x) =x^2({(2x-5)}^3)(8 x) + 3(2x-5)) </math>
<math>a'(x) =x^2({(2x-5)}^3)(8 x) + 3(2x-5)) </math>


<math>a'(x) =x^2({(2x-5)}^3)(8 x+6x-15)) </math>
<math>a'(x) =x^2({(2x-5)}^3)(8 x+6x-15)) </math>
<math>a'(x) =x^2{(2x-5)}^3(14x-15)) </math>
<math>a'(x) =x^2{(2x-5)}^3(14x-15)) </math>
====example 2====
====example 2====
<math>f'(x)= x^2 \sqrt[2]{4-9x}</math> simplify  
<math>f'(x)= x^2 \sqrt[2]{4-9x}</math> simplify  
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<math> { x (8 - { 45x \over 2} ) \sqrt[2]{4-9x} \over \sqrt[2]{4-9x} }</math>
<math> { x (8 - { 45x \over 2} ) \sqrt[2]{4-9x} \over \sqrt[2]{4-9x} }</math>
====example 3====
<math>f(x)= x^2 \sqrt[2]{1-x^2}</math>  simplify
<math>f(x)= x^2 {(1-x^2)}^{1 \over 2}</math>
Product Rule <math>(f*g)'=f*g'+ g*f'</math>
<math>f'(x)= x^2 ( 1 \over 2 {(1-x^2)}^{-1 \over 2} (-2x) ) + 2x {(1-x^2)}^{1 \over 2}</math>
<math>f'(x)= -x^3 {(1-x^2)}^{-1 \over 2} + 2x {(1-x^2)}^{1 \over 2}</math>
factorize
<math>f'(x)= x {(1-x^2)}^{-1 \over 2} [-x^2 + 2 {(1-x^2)}]</math>
<math>f'(x)= {x \over {(1-x^2)}^{1 \over 2}} [{-x^2 + 2 -2x^2}]</math>
<math>f'(x)= {x \over {(1-x^2)}^{1 \over 2}} {[ 2 - 3x^2]}</math>
<math>f'(x)= {x [ 2 - 3x^2] \over {(1-x^2)}^{1 \over 2}} </math>
<math>f'(x)= {[ 2x - 3x^3] \over {(1-x^2)}^{1 \over 2}} </math>
====example 4====
<math>f(x)= {(x+3)^4 \over (x^2 + 5)^{1 \over 2} }</math>
The Quotient Rule <math>({f \over g})' = {(gf'-fg') \over g^2}</math>
<math>f = {(x+3)}^4 </math>
<math>f' = 4(x+3)^3 </math>
<math>g = (x^2 + 5)^{1 \over 2} </math>
<math>g' = {1 \over 2} (x^2 + 5)^{-1 \over 2} (2x) </math>
<math>{{(x^2 + 5)^{1 \over 2} 4(x+3)^3 + {(x+3)}^4 {1 \over 2} (x^2 + 5)^{-1 \over 2} (2x) } \over (x^2 + 5) }</math>
<math>{{4{(x^2 + 5)}(x+3)^3 - x{(x+3)}^4 (x^2 + 5)^{-1 \over 2} } \over (x^2 + 5) }</math>
<math>(x^2 + 5)^{-1 \over 2} {(x+3)}^3 [4(x^2 + 5)- x(x+3)] \over (x^2 +5)</math>
<math> {(x+3)}^3 [4(x^2 + 5)- x(x+3)] \over (x^2 +5) (x^2 + 5)^{1 \over 2}</math>
<math> {(x + 3)}^3 [4x^2 + 20 - x^2 - 3x] \over (x^2 +5) (x^2 + 5)^{1 \over 2}</math>
<math> {(x + 3)}^3 [3x^2 - 3x + 20] \over (x^2 + 5)^{3 \over 2}</math>
<math> {(x + 3)}^2 [3x^2 - 3x + 20] \over (x^2 + 5)^{1 \over 2}</math>

Latest revision as of 12:12, 10 October 2021

example 1

Product Rule


factorize

example 2

simplify

Chain rule

example 3

simplify

Product Rule

factorize



example 4

The Quotient Rule