Question

Derive y= 1/(5^2-x^2)

Why does my calculator say the solution is 2x/(x^2-25)

When my worked out solution is 2x/(25-x^2)?

Answer 1

\[y=\frac{1}{25-x^2}~~\Rightarrow~~y'=-\frac{(-2x)}{(25-x^2)^2}=\frac{2x}{(25-x^2)^2}=\frac{2x}{(x^2-25)^2}\]
Recall that
\[\color{red}{(a-b)^2}=((-1)(b-a))^2=(-1)^2(b-a)^2=\color{red}{(b-a)^2}\]

Answer 2

Thank you much.
It makes complete sense.

Then the proper way is to have a positive leading term, which is why the calc came out that way.

But both solutions are correct because the denominator is squared anyways (which squares i failed type in)

Thanks again

^_^