Asked by Robin
What would an example of a logarithmic function that contains a radical within it and a quotient within it where you would need to use chain rule and quotient rule to take derivative look like? How would taking the derivative of the function in its original form look different from use log properties to simplify the function first and then take derivative?
Answers
Answered by
Reiny
did you mean something like
y = √( (x^2 + 5x)/(5x) )
original form derivative:
dy/dx= (1/2)((x^2+5)/(5x))^(-1/2) ( 5x(2x) - (x^2+5)(5))/(25x^2)
etc , but still has to be simplified
using logs first:
take ln of both sides:
ln y = ln((x^+5)/(5x))^(1/2)
= (1/2)(ln (x^2+5) - ln (5x) )
(dy/dx)/y = (1/2)( 2x/(x^2+5) - 5/(5x) )
= x/(x^2 + 5) - x/2
dy/dx = y( x/(x^2 + 5) - x/2 )
or
= (√( (x^2 + 5x)/(5x) )( x/(x^2 + 5) - x/2 )
this may not look any simpler, but it was easier to get to.
y = √( (x^2 + 5x)/(5x) )
original form derivative:
dy/dx= (1/2)((x^2+5)/(5x))^(-1/2) ( 5x(2x) - (x^2+5)(5))/(25x^2)
etc , but still has to be simplified
using logs first:
take ln of both sides:
ln y = ln((x^+5)/(5x))^(1/2)
= (1/2)(ln (x^2+5) - ln (5x) )
(dy/dx)/y = (1/2)( 2x/(x^2+5) - 5/(5x) )
= x/(x^2 + 5) - x/2
dy/dx = y( x/(x^2 + 5) - x/2 )
or
= (√( (x^2 + 5x)/(5x) )( x/(x^2 + 5) - x/2 )
this may not look any simpler, but it was easier to get to.
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