Question

find f '' (x) from : f(x)=(3x -2)^5

Answer 1

did you find the first derivative?

Answer 2

so.. how to find the first derivative...?

Answer 3

Use the chain rule

Answer 4

f(g(x))' = f'(g(x))g'(x)

Answer 5

\[f'(x) = 15\, \left( 3\,x-2 \right) ^{4}\]\[f''(x) = 180\, \left( 3\,x-2 \right) ^{3}\]

Answer 6

maybe you can explain in detail.?

Answer 7

You differentiate the function twice.

Answer 8

Are you familiar with the chain rule?

Answer 9

no...

Answer 10

I hope u can describe for me...

Answer 11

You should learn a bit about differentiation first before you can work on these problems.

Answer 12

I have learn about differentiation first may be like this :
f (x) = 3x (x ^ 3-1) =3x^4 -3x f'(x)=12x-3

Answer 13

Think of the chain rule as layes on a cake

Answer 14

The function you have is a cake

Answer 15

The first layer is x^5, the second layer is the inner function or 3x-2

Answer 16

ok... I have started to understand...

Answer 17

Now, with the chain rule, you differntiate the first layer leaving the inner layer untouched

Answer 18

Then differntiatie just the inner layer

Answer 19

fiest u take 3x-2 as y
then differentiate y^5 to get 5y^4
then using chain rule differentiate 3x-2
to get 3(5)y^4..
similarly for f''(x)

Answer 20

\[f(x) = x^5\]\[g(x) = 3x-2\]\[(f(g(x))' = f'(g(x))g'(x)\]\[f'(x) = 5x^4\]\[g'(x) = 3\]\[f'(g(x))g'(x) = 5(3x-2)^4 \cdot 3 = 15 \cdot ( 3x-2)^4\]

Answer 21

thanks Alchemista...