Question

i need to find the dy/du, du/dx, and dy/dx when u=7x-2x^2 and y - u^(1/2)

Answer 1

Did you mean u=7x-2x^2 and y = u^(1/2) ?

If so, then find the derivatives of those. Then using chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}\]
you already have du/dx and dy/du

Answer 2

Alright basically dy/dx just means derivative of y (function) in respect to x(for values of x)

Answer 3

so for dy/du you have

\[y=u^{\frac{1}{2}}\]
solving this by using your chain rule you'll get
\[\frac{dy}{du}=\frac{1}{2}u^\frac{-1}{2}(1)\]

Answer 4

next: du/dx

\[\frac{du}{dx}=7-4x\]

Answer 5

lastly dy/dx

since you know the derivatives of the two you can multiply them together like IanT said which will cancel the du and get dy/dx or you can substitute u=7x-2x^2 into your y equation to get

\[y=7x-2x^2\]

Answer 6

my mistake you'll get
\[y=(7x-2x^2)^\frac{1}{2}\]

Answer 7

using the multiplication way or this way you'll get\[\frac{dy}{dx}=\frac{1}{2}(7x-2x^2)^{\frac{-1}{2}}(7-4x)\]

Answer 8

simplifying you get
\[\frac{7-4x}{2\sqrt{7x-2x^2}}\]