Compute the derivative dy/dx and simplify your answer.
y = u^2 + u - 2
u = 1/x or x^1/2
y' = 2u + 1
u' = 1/2(x)^-1/2 = 1/2(square root x)
2(1/x) + 1 *1/2(square root x)
2/x + 1 * 1/2(square root x)
$\displaystyle y'=2u+1$Originally Posted by startingover
$\displaystyle u=1/(2\sqrt{x})$
Thus,
$\displaystyle 2(1/2(\sqrt{x}))+1=1/\sqrt{x}+1$
Multiply,
$\displaystyle 1/(2\sqrt{x})(1/\sqrt{x}+1)$
Thus,
$\displaystyle 1/(2x)+1/(2\sqrt{x})$
Not trying to be a pain - honest...
In the beginning equation, why did you substitute 1/2square rt x in for the u expression? Shouldn't it be the orginal u, not the derivative of u?
y = u^2 + u - 2
y' = 2u + 1
u = 1/x
u' = 1/2squarert x
2 (1/x) + 1 * 1/2squarert x
If $\displaystyle u = \frac{1}{x}$ then $\displaystyle u \ne x^{\frac{1}{2}}$Originally Posted by startingover
$\displaystyle u = x^{-1}$
$\displaystyle u' = -x^{-2}$
Have $\displaystyle \frac{dy}{du} = 2u + 1$ and $\displaystyle \frac{du}{dx} = -x^{-2}$
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
$\displaystyle = (2u + 1)(-x^{-2})$
$\displaystyle = (2x^{-1} + 1)(-x^{-2})$
$\displaystyle = -2x^{-3} - x^{-2}$
$\displaystyle = -\frac{2}{x^3} -\frac{1}{x^2} = -\frac{2 + x}{x^3}$
Using the ' notation for derivative here is confusing as it means d/dx in oneOriginally Posted by startingover
place and d/dy in another.
The chain rule is:
dy/dx=dy/du du/dx
so:
dy/du=2u+1
and:
du/dx=-1/x^2 (that is when u=1/x)
So:
dy/dx=-(2u+1)/x^2=-(2/x+1)/x^2=-(x+2)/x^3
RonL