Mathematics: Bored anyone wanna talk? or do a zo om?

Bored anyone wanna talk? or do a zo om?

legendman27

17.11.2021

$(G(s(z)))'=G'(s(z))\cdot s'(z))$

Step-by-step explanation:

If G is a function of s, and s is a function of z, then the composition function is :

$(G\circ s)(z)=G(s(z))$

This is a function of a function. So we apply the chain rule to different the outer function multiply by the derivative of the inner function.

We take the first derivative to obtain:

$(G(s(z)))'=G'(s(z))\cdot s'(z))$

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