Find dy/du, du/dx, dy/dx, when y and u are the following

Question

when Y=4u^2 + 2 & u= x^2 + 3

what is

dy/du=

du/dx=

dy/dx=

Answer 1

Hello Ernesto!

I'll try and help you out with this problem!

Starting Equations:

y = 4u² + 2

u = x² + 3

To find dy/du and du/dx we don't need to do anything out of the ordinary, just take the derivatives of our two starting equations like so:

y = 4u² + 2

dy/du = 8u

and...

u = x² + 3

du/dx = 2x

But, for dy/dx we need to take a different approach, since we do not have an expression for y in terms of x.

Imagine treating these differentials kind of like how we multiply fractions. Let's multiply them like so and see what happens...

[dy/du] * [du/dx] = (dy*du) / (du*dx)

We can combine the differentials just like how we multiply fractions. Now notice how there is a du term in both the numerator and the denominator, therefore we can cancel them both out!

[dy/du] * [du/dx] = (dy*du) / (du*dx) = (dy/dx)

This expression is just equal to the differential we wanted to solve! Let's plug in our known values for [dy/du] and [du/dx] ...

(dy/dx) = [dy/du] * [du/dx]

(dy/dx) = [ 8u ] * [ 2x ]

(dy/dx) = 16ux

If we want an answer only in terms of x, then we can plug our expression for u back into this formula...

(dy/dx) = 16ux ; u = x² + 3

(dy/dx) = 16*(x² + 3)*x

(dy/dx) = 16*(x³ + 3x)

(dy/dx) = 16x³ + 48x

Alternatively, you can solve for dy/dx by plugging our equation for u in terms of x into our equation for y in terms of u -- that gives you an equation for y in terms of x that you can then differentiate for dy/dx.

Hope that helps!

--Zach