D Dr_fox New member Joined Feb 3, 2021 Messages 4 Monday at 2:05 PM #1 d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R
d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R
S Subhotosh Khan Super Moderator Staff member Joined Jun 18, 2007 Messages 23,789 Monday at 2:21 PM #2 Dr_fox said: d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R Click to expand... Did you have a question? Please write out the result of your work in mathematics, instead of using only words.
Dr_fox said: d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R Click to expand... Did you have a question? Please write out the result of your work in mathematics, instead of using only words.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,176 Tuesday at 8:30 AM #3 Dr_fox said: d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R Click to expand... FIRST use the product rule, then use the chain rule. Also you say that "R is a function of t" but repeatedly write "R(I, t)", making it look like R is a function of both I and t. It isn't- R^2 is simply multiplied by I. \(\displaystyle \frac{dIR^2}{dt}=\frac{dI}{dt}R^2+ I\frac{dR^2}{dt}\) \(\displaystyle = \frac{dI}{dt}R^2+2IR\frac{dR}{dt}\).
Dr_fox said: d/dt(IR^2) first i use the chain rule and got 2R(I,t)d/dt(R(I,t) then took the derivative of R(I,t) wich is I2R(I,t)d/dt(R Click to expand... FIRST use the product rule, then use the chain rule. Also you say that "R is a function of t" but repeatedly write "R(I, t)", making it look like R is a function of both I and t. It isn't- R^2 is simply multiplied by I. \(\displaystyle \frac{dIR^2}{dt}=\frac{dI}{dt}R^2+ I\frac{dR^2}{dt}\) \(\displaystyle = \frac{dI}{dt}R^2+2IR\frac{dR}{dt}\).
