Question

best way for find derivate of this:
d/ds(z/((s+1)(s+2)(s+3)^2)(z-e^TS))
please help me i need fastest way for solve this or problem like this
thank you

Answer 1

i find that this is chain Rule

Answer 2

This is what you have to calculate:\[\frac{ d }{ ds }\left( \frac{ z }{ (s+1)(s+2)(s+3)^2(z-e^{ts}) } \right)\]Because s is the variable, and z and t are constants, we also could write it as:\[z \frac{ d }{ ds }((s+1)(s+2)(s+3)(z-e^{ts}))^{-1}=\]\[-z((s+1)(s+2)(s+3)(z-e^{ts}))^{-2} \cdot \frac{ d }{ ds }((s+1)(s+2)(s+3)(z-e^{ts}))\]So it is the Chain Rule indeed, but as you can see, you also need the Product Rule.
I think it is more convenient now to calculate the second part separately:\[\frac{ d }{ ds }((s+1)(s+2)(s+3)(z-e^{ts}))=\]\[(s+2)(s+3)(z-e^{ts})+\]\[(s+1)(s+3)(z-e^{ts})+\]\[(s+1)(s+2)(z-e^{ts})+\]\[(s+1)(s+2)(s+3)(-te^{ts})\]

All you have to do now is write everything as one fraction and you're done.
Seems a lot of work, btw....

Answer 3

Oh, I see it's even worse: I've lost the square in (x+3)^2 :(.

Are you really sure of this rather weird function?

Answer 4

assuming z iz constant, the easiest way is to use logarithmic differentiation

Answer 5

@watchmath : could you show how?

Answer 6

let say that your big expression is x. Then \(\ln x= \ln z - \ln\left( (s+1)(s+2)(s+3)^2(z-e^{ts})\right)\). Use the log property to write the second terms on the right hand as the sum of logs. Then take the derivative implicitly with respect to s

Answer 7

I see! That reduces the calculations by half a page ;)