80s Male Aerobic Instructor

Derivata di una funzione di funzione

This is' perhaps the 'do more' important to know how to exactly calculate the derivative: To make the derivative of a function of the derivative of the function before I external function without touching the inside and then multiply by the derivative of the inside.
In symbols, if I
y = f (g (x))

then y '= f' (g (x)) · g '(x)
Let's understand this with an example
y = sin (logx )
I do the first derivative of the breast and that 'cos
then the first part of the derivative of
sen (logx) will' cos (logx)
as if we had instead of x logx
now I have to do and that the derivative of logx '1 / x
then I'll have' y '= cos (logx) • 1 / x -----------
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To make it 'easy to think of an onion: the onion and' made in layers to peel and I have to remove the first layer, then the second , then the third ...
The role and function of 'layered, first I have to derive the first function and leave the other, then the second .... I have left until last when the x
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we see another example;
y = ( log (senx) 5 Here I
exponentiation function 5 which contains the log that contains the breast surrounding the root that contains x
Before I do the derivative of the power 5: x5
if the derivative is 5x4, in this case because 'instead of x I log (senx) the first part of the derivative will be '
5 (log (senx) 4
turn now to the second function and that' the logarithm:
logx if the derivative is 1 / x,
because 'instead of x I senx
the second part of the derivative will be ':
1 / (senx)
turn now to the third function and that' the breast
senx if the derivative is cosx,
because 'instead of x I x
the third part of will be derived ':
cosx
Step hours and the fourth function that 'the root
the derivative of x' 1 / (2x) and I came to this and then x 'the last part
collecting
y' = 5 (log (senx) 4 ° [1 / (senx)], cosx · [1 / (2x)]