> coef(graphmodeld)
> (Intercept) graphdata[, 1]
> 4.513544204 -0.006820623
>
> This corresponds to the linear model but it is still not understood how this
> helps
> to determine gradients at specific points for the exponential model.
You have fitted log(y) against x. Your fitted linear model is therefore
log(y) = 4.5135 - 0.00682x
>From there:
- do the basic algebra to get y in terms of x
- differentiate analytically to get dy/dx as a function of x.
- write that calculation in R to get your gradients at any value of x.
Or you could get R to give you the differential function from your function for
y; see ?D
S Ellison
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