Chain rule with leibniz notation

[[ScPpL]Shree]

Homework Statement

If y=f((x²+9)^0.5) and f'(5)=-2, find dy/dx when x=4

Homework Equations

chain rule: dy/dx=(dy/du)(du/dx)

The Attempt at a Solution

In my opinion giving f'(5)=-2 is unnecessary as:

y=f(u)=u, u=(x²+9)^0.5

dy/dx= (dy/du)(du/dx)

(dy/du)= 1
(du/dx)= x/((x²+9))^0.5

dy/dx = (1)(x/((x²+9))^0.5)
= x/((x²+9))^0.5
dy/dx(4) = 4/(16+9)^0.5
= 4/(25)^0.5
= 4/5

the answer is -8/5

I would appreciate help very much.

 

[annoymage]

wait, how do you get f(u)=u?

 

[Mark44]

Homework Statement

If y=f((x²+9)^0.5) and f'(5)=-2, find dy/dx when x=4

Homework Equations

chain rule: dy/dx=(dy/du)(du/dx)

The Attempt at a Solution

In my opinion giving f'(5)=-2 is unnecessary as:

y=f(u)=u, u=(x²+9)^0.5

As Annoymage pointed out, you can't assume that f(u) = u.

dy/dx= (dy/du)(du/dx)

(dy/du)= 1

y = f(u), so dy/dy = f'(u)

(du/dx)= x/((x²+9))^0.5

dy/dx = (1)(x/((x²+9))^0.5)
= x/((x²+9))^0.5
dy/dx(4) = 4/(16+9)^0.5
= 4/(25)^0.5
= 4/5

the answer is -8/5

I would appreciate help very much.