Finding the Derivative of y= (x^2-x)^1/2 (x-1)

[Stratosphere]

Homework Statement

Find the derivative of $$\ y=\sqrt{x^{2}-x}(x-1)$$

Homework Equations

The Attempt at a Solution

I'm not sure how to begin for this one. All I know to do is to change $$\sqrt{x^{2}-x}$$ to (x^2-x)^1/2 Do you I have to take the derivative of each one separately?

 

[rock.freak667]

Firstly, you will see that y is of the form y=uv so you will need your product law here.

so u=(x²-x)^1/2 right?

Now you want to get du/dx

so the chain rule must be used now. if you put t=x²-x, then what is dt/dx equal to? Then what is u equal to in terms of t?

 

[Stratosphere]

Using the product rule I got, 1/2(x^2-x)$$^{-1/2}$$(x-1)+(x$$^{2}$$-x)$$^{1/2}$$. that doesn't seem right, did I mess somthing up?

 

[rock.freak667]

Using the product rule I got, 1/2(x^2-x)$$^{-1/2}$$(x-1)+(x$$^{2}$$-x)$$^{1/2}$$. that doesn't seem right, did I mess somthing up?

I made a typo that probably confused you.

So I will start over.

$$y=\sqrt{x^{2}-x}(x-1)$$

so u= (x²-x)^1/2
and v= x-1 (dv/dx is easily found right?)

The problem lies with du/dx...so we let t=x²-x and so u = ?? (in terms of t)

to apply the chain rule now

$$\frac{du}{dx}= \frac{d?}{dx} \times \frac{d??}{dt}$$

can you think of what variable '?' is and '??' is? (hint: 'd?' will cancel out with dt)

 

[Stratosphere]

dv/dx=1 but what is t?

 

[rock.freak667]

dv/dx=1 but what is t?

yes

t is a new variable we introduced for u=(x²-x)^1/2

we said t =x²-x , so u=t^1/2

from t=x²-x, what derivative can we find?

From u=t^1/2 what derivative can we find?

 

[Stratosphere]

$$\ \frac{dt}{dx}=2x-1$$

$$\ \frac{du}{dt}=1/2t^{-1/2}$$

So by multiply them together I will find dy/dx?

 

[rock.freak667]

$$\ \frac{dt}{dx}=2x-1$$

$$\ \frac{du}{dt}=1/2t^{-1/2}$$

So by multiply them together I will find dy/dx?

you will find du/dx.


You used the product rule correctly in post#3, but du/dx was wrong.

 

[Mentallic]

I remember when I began learning to take derivatives using the product rule, it never occurred to me to use other variables such as t when finding dy/dx, and I did just fine without it too. Only once I was very familiar with it all did I start using other variables to apply the function of a function rule.

Stratosphere, maybe this is the best approach for you to take as well?

Since $$y=\sqrt{x^2-x}(x-1)$$ requires the product rule:

$$y=uv$$ then $$y'=u'v+v'u$$ where $$u=\sqrt{x^2-x}$$ and $$v=x-1$$

Just take each variable in the product rule separately:

first, what is $$u'$$? $$u'=\frac{d}{dx}(\sqrt{x^2-x})$$
next, what is $$v'$$?...
etc.
Once you have them all, just substitute into the product rule formula.

Remember that if $$y=[f(x)]^n$$ then $$\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)$$
When you tried applying the product rule earlier, you neglected the last $$f'(x)$$ bit.

 

[djeitnstine]

Hmm I propose this method $$\sqrt{x(x-1)}(x-1)$$

$$\sqrt{x}(x-1)^{\frac{1}{2}}(x-1)^1$$ by the power laws we get

$$x^{\frac{1}{2}}(x-1)^{\frac{3}{2}}$$ I believe this seems less messy to deal with.

 

[Mentallic]

Hmm I propose this method $$\sqrt{\frac{1}{x}(x-1)}(x-1)$$

$$\frac{1}{\sqrt{x}}(x-1)^{\frac{1}{2}}(x-1)^1$$ by the power laws we get

$$x^{-\frac{1}{2}}(x-1)^{\frac{3}{2}}$$ I believe this seems less messy to deal with.

You made a typo. $$\sqrt{x^2-x}=\sqrt{x(x-1)}\neq\sqrt{\frac{1}{x}(x-1)}$$

and yes I agree with you, it does make the question simpler.

 

[djeitnstine]

Yes I fixed it thanks