When am i suppose to use this rule when im diffrentiating

[pliant canyon]

so

I know the normal rules

this and

[frosty citrus]

if you have a composite of functions

so something like sin(6x)

requires the chain rule

[pliant canyon]

so

cos(5x)

is then

-sin(5x) x 5

or am i wrong

[frosty citrus]

yes
or -5sin(5x)

but i guess your formatting is essentially the same

[pliant canyon]

but

I dont understand

how

composite functions mean

I know how to use it

but dont know when

like when the sin (5x) is between brakkets

[frosty citrus]

if you have a function that's inside of another function then you use the chain rule

[pliant canyon]

so you don't use the chainrule when it is like f(x)= sin(x)

if there is somthing else besides the x then you use the rule?

between the brackets

[frosty citrus]

you technically do
but if you define sin(x) as (u) and (x) as v, then u' = cos(x) and v' = 1, so now you have:

1 • cos(x)

which is just cos(x)

[pliant canyon]

yeah

so you use it only if there is somthing else beside x

otherwise you just multiply by 1

[frosty citrus]

yes

since the derivative of x is 1

[pliant canyon]

yeah

can you give me an example

other then sin cos tan

[frosty citrus]

ok

[pliant canyon]

✅

I don't know if you were going to send the example question but i think i get it thank you

[frosty citrus]

i am

wait

[pliant canyon]

oh ok

[frosty citrus]

f(x) = (3x^2 + 2x)^2

f'(x):
we can define variables, the inner part being (u), and the outer part being (v)

u = 3x^2 + 2x
v = 1(u)^2

u' = 6x + 2
v' = 2u

substitute u back in and apply the chain rule, where (IN THIS CASE) it's given as:

v'(u(x)) • u'(x)

remember that:
v = 1(u)^2
u = 3x^2 + 2x
v' = 2u
u' = 6x + 2

2(3x^2 + 2x)(6x + 2)
↓
(6x^2 + 4x)(6x + 2)
which simplifies to 36x^3 + 36x^2 + 8x
f'(x) = 36x^3 + 36x^2 + 8x

@pliant canyon um do you understand

[pliant canyon]

im reading

yes thankyou

@frosty citrus

i get it

100

thankyou sir