What are the steps to simplify the following equation (with simplified version included)

Question

This is part of a question asking to find the differential of a polynomial fraction. I have already taken the derivative using quotient and chain rule, and this is where I am up to.

Rather than go through the entire question (I have been told this is too confusing for readers in previous questions), I will simply show the area where I am stuck.

I need to know how to simplify : $$\frac{\sqrt{2x-x^2}-\frac{-x+1}{\sqrt{2x-x^2}}x}{(\sqrt{2x-x^2})^2}$$

to : $$\frac{x}{(2x-x^2)^{3/2}}$$

Any help with this would be very much appreciated , exam is tomorrow .

P.S. If users would prefer me to write out the full question, just ask and I will. I am still learning to write the perfect question.

Answer 1

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

Can you continue from here?

Answer 2

Look at it as $(\sqrt A-(-x+1)x/\sqrt A)/(\sqrt A)^2$ with $A=2 x-x^2.$

Assuming $A> 0$ we have $(\sqrt A)^2=A=2 x-x^2,$ which,so far,is a simpler denominator. (Actually the notation $\sqrt A$ should only be used when $A\geq 0.$ And we must have $A\ne 0$ else we have a denominator $0$.)

Now get rid of that pesky minus-sign between the two terms of the numerator by re-writing the numerator as $\sqrt A+(x-1)x/\sqrt A$ and re-write this as $\sqrt A+(x^2-x)/\sqrt A .$

The big step now is to multiply both the numerator and denominator by $\sqrt A$ to remove the occurrence of a denominator $\sqrt A$ within the numerator of the whole expression. This gives us $((\sqrt A)^2+\;x^2-x)/ (A\sqrt A).$

Once again we have $(\sqrt A)^2=A$ so we can write the numerator as $A+(x^2-x).$ But what is $A$? It's $2 x-x^2.$ So the numerator is $(2 x-x^2)+(x^2-x).$ Now look for a re-grouping or for terms that cancel out: $2 x-x^2+x^2-x=2 x-x=x $ is now the numerator.

So we obtain $x/(A\sqrt A)=$ $x/(A^{3/2})=$ $x/(2 x-x^2)^{3/2}.$

We can also at this point divide both the numerator and denominator by $\sqrt x$ to obtain $\sqrt x/(2-x)^{3/2}$ and, if we like, write this as $(x/(2-x)^3)^{1/2}$.