Differentiation Question
Hi, could someone please explain this to me. In this question, I know N=900/3+...., and I have dN/dt in terms of N, so if I wanted to differentiate dN/dt is it ok to just differentiate with respect to N and sub back in the value of N? Or do I need to also do something with the value of N? I know in chain rule you differentiate the value of u and then multiply this with dy/du to get dy/dx, but this seems different. Thanks. (this is for part (c), I do know that you can just use the symmetricallity of the dN/dt curve to find N=150 but I just wanted to make sure I fully understood).
(edited 12 months ago)
Its, dN/dt, a quadratic in N, so find (write down) that value of N which maximises it and rearrange N(t) for t?
You could also differentiate dN/dt with respect to N (and set equal to zero) as its a function of N. Just use the chain rule for the derivative of dN/dt with respect to t to understand why, so
d/dt dN/dt = d/dN dN/dt * dN/dt
so the derivative of the gradient is zero (gradient stationary point) if either of the two derivatives on the right are zero. dN/dt is an increasing function, so it must be the first term thats zero.
You could also differentiate dN/dt with respect to N (and set equal to zero) as its a function of N. Just use the chain rule for the derivative of dN/dt with respect to t to understand why, so
d/dt dN/dt = d/dN dN/dt * dN/dt
so the derivative of the gradient is zero (gradient stationary point) if either of the two derivatives on the right are zero. dN/dt is an increasing function, so it must be the first term thats zero.
(edited 12 months ago)
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