Given a function, finds its derivative?

Question

y= (2x-3)^7

u= 2x-3

y= u^7

Find the derivatives of the simpler functions

d/dx= (2x-3)2 Where did the 2 come from at the end?

d/du= (u^7)

Apply the chain rule and simplify

dy/du x du/dx

=(7u^6)(2)
=14u^6
Am I missing some notes here becauce jupming from here to here doesn't make sense. The (2x-3) just disappears?
= 14 (2x-3)^6

Can't I just use the power rule and get this faster?

x^p = x^(p-1)

So

7 x (2x-3)^(7-1)

but that equals 14x+21^6 so I guess I can't use that right?

Answer 1

y = (2x - 3)^7

The derivative of (2x -3) is 2 -- that is where the 2 came from.

Since it is raised to the 7th power, the derivative is multiplied by 7 and the power decreases to 6, so

dy/dx = 7*(2x-3)^6 *2
= 14 *(2x-3)^6

Now the way you have in your notes is to find the derivate by substitution. So:
y = (2x-3)^7
u = 2x - 3
du/dx = 2

y = u^7
dy/du =7 u^6

Then
dy/dx = dy/du * du/dx
dy/dx = 7 u^6 * 2
dy/dx = 14 u^6

But u = 2x-3, so substiting this back in, we get:

dy/dx = 14 * (2x-3)^6

I hope this helps!

Answer 2

Where did the 2 come from at the end?
Chain Rule - derivative of 2x - 3 is 2.

The 2x - 3 didn't disappear, it was disguised as u.

I would just apply the chain rule to the original:
y = (2x - 3) ^7

y' = 7(2x - 3)^6 * 2

y' = 14(2x - 3)^6

Answer 3

y = (2x-3)^7

u = 2x-3 -----> du= 2 dx ( 2 is the derivative of 2x-3)
y = u^7
dy= 7u^6 du

substitute,
dy = 7 [(2x-3)^6] 2 dx
=7(2) (2x-3)^6 dx
=14 (2x-3)^6 dx
dy/dx = 14 (2x-3)^6

Answer 4

Let u = 2x - 3
du/dx = 2
y = u^7
dy/du = 7u^6
dy/dx = dy/du X du/dx
dy/dx = 7 u^6 X 2
dy/dx = 14.(2x - 3)^6