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.

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- May 1st 2008, 06:25 PMwhite_capEasy 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. - May 1st 2008, 06:50 PMMr. Engineer

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 $\displaystyle \frac{dy}{dx}$ - May 1st 2008, 06:52 PMo_O
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