Math Help - Help with a few basic derivatives...

1. Help with a few basic derivatives...

1. y = (x^2 + 1)^3 * (x^3 - 1)^2

2. y = (x^3 + 1) / (x^3 - 2)

3. y = x^2 * sqrt(x + 1)

4. find dy/dx and d^2y/dx^2 for x^(1/3) + y^(1/3) = 3 at the point P(1,8)

If you could briefly explain the methods that would be great too. These are the only ones I am stuck on...thanks1

2. Originally Posted by golfman44
1. y = (x^2 + 1)^3 * (x^3 - 1)^2

2. y = (x^3 + 1) / (x^3 - 2)

3. y = x^2 * sqrt(x + 1)

4. find dy/dx and d^2y/dx^2 for x^(1/3) + y^(1/3) = 3 at the point P(1,8)

If you could briefly explain the methods that would be great too. These are the only ones I am stuck on...thanks1

$(uv)'=u'v+uv'\,\,\,and\,\,\,\left(\frac{u}{v}\righ t)'=\frac{u'v-uv'}{v^2}$ for any two differentiable functions, so for example

$\frac{x^3+1}{x^3-2}=1+\frac{3}{x^3-2}\,\,\,so\,\,\left(1+\frac{3}{x^3-2}\right)'=\frac{-9x^2}{\left(x^3-2\right)^2}$

About the last one: differentiate implicitly every summand wrt x and y:

$\frac{1}{3}x^{-\frac{2}{3}}dx+ \frac{1}{3}y^{-\frac{2}{3}}dy=0\Longrightarrow \frac{dy}{dx}=\left(\frac{y}{x}\right)^{\frac{2}{3 }}$ and now plug in (1,8)

For the second derivative $\frac{dy^2}{dx^2}$ do as above with every summand.

Tonio

3. Originally Posted by tonio

About the last one: differentiate implicitly every summand wrt x and y:

$\frac{1}{3}x^{-\frac{2}{3}}dx+ \frac{1}{3}y^{-\frac{2}{3}}dy=0\Longrightarrow \frac{dy}{dx}=\left(\frac{y}{x}\right)^{\frac{2}{3 }}$ and now plug in (1,8)

For the second derivative $\frac{dy^2}{dx^2}$ do as above with every summand.

Tonio
I think Tonio made a small mistake: the derivative for this last one should be $\frac{dy}{dx} = - (\frac{y}{x})^{\frac{2}{3}}$

Because this is actually a function, there is no need to differentiate implicitly, as the equation can be rearranged to find y explicitly in terms of y:

$y = (3 - x^{\frac{1}{3}})^3 = 27 - 27x^{\frac{1}{3}} + 9x^{\frac{2}{3}} - x$

Then: $\frac{dy}{dx} = -9x^{\frac{-2}{3}} + 6x^{\frac{-1}{3}} - 1$. Now just sub in x = 1.

For $\frac{d^2 y}{dx^2}$: $\frac{d^2 y}{dx^2} = 6x^{\frac{-5}{3}} - 2x^{\frac{-4}{3}}$. Again sub in x=1.