Difference between revisions of "Simplifying Derivatives"
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<math>f'(x)= x {(1-x^2)}^{-1 \over 2} [-x^2 + 2 {(1-x^2)}]</math> | <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}} [{-x^2 + 2 -2x^2}]</math> | ||
<math>f'(x)= {x \over {(1-x^2)}^{1 \over 2}} [ 2 - 3x^2} | <math>f'(x)= {x \over {(1-x^2)}^{1 \over 2}} {[ 2 - 3x^2]}</math> | ||
<math>f'(x)= {x [ 2 - 3x^2 | <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>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