# Thread: Easy question: i'm new too calc. (urgent)

1. ## Easy question: i'm new too calc. (urgent)

y= u^5 +u^3

u= 3/v - 4v

v= 3 - x^2

Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

Text gives the solution: -48608

do i use:

(dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer.

2. Originally Posted by white_cap
y= u^5 +u^3

u= 3/v - 4v

v= 3 - x^2

Use the chain rule in Leibniz notation to find (dy/dx) when x = 2

Text gives the solution: -48608

do i use:

(dy/dx) = (dy/du)(du/dv)(dv/dx) because when I do, I don't get the right answer.

you do use the chain rule, but first you have to substitute to get your function in terms of x
$\displaystyle u=\frac{3}{3-x^2}-4(3-x^2)$
then distrubute
$\displaystyle u=\frac{3}{3-x^2}-(12-4x^2)$

y= u^5 +u^3
$\displaystyle y=(\frac{3}{3-x^2}-(12-4x^2))^5+(\frac{3}{3-x^2}-(12-4x^2))^3$
now u can differentiate this function using the chain rule in terms of $\displaystyle \frac{dy}{dx}$

3. Well take the derivative:
$\displaystyle \frac{dy}{dx} = {\color{blue}\frac{dy}{du}} \cdot {\color{red}\frac{du}{dv}} \cdot {\color{magenta}\frac{dv}{dx}}$
$\displaystyle \frac{dy}{dx} = {\color{blue} \left(5u^{4} + 3u^{4}\right) } \cdot {\color{red} \left(-3v^{-2} - 4\right)} \cdot {\color{magenta}-2x}$ (Just the application of the power rule over and over again)

Now use x = 2 to find v(2) and u(2):
$\displaystyle v(2) = 3 - (2)^{2}$
$\displaystyle u(2) = \frac{3}{v(2)} - 4v(2)$

Then plug all these values into your expression for $\displaystyle \frac{dy}{dx}$ at x = 2