I think you may be misunderstanding what your teacher is saying. If you were getting a "common denominator" to add or subtract fractions, say, then you would need to multiply both numerator and denominator by the same thing so you get a different form of the same fraction.
If you have the fraction, say, \(\displaystyle \frac{3}{4}\), and want to add it to \(\displaystyle \frac{2}{5}\), you want to multiply both numerator and denominator by 5 so that you have \(\displaystyle \frac{15}{20}\) while multiplying both numerator and denominator of \(\displaystyle \frac{2}{5}\) by 4 to get \(\displaystyle \frac{8}{20}\). Now they both have denominator 20 and so can be added. But you have only changed the "form" of the fractions, not the value-\(\displaystyle \frac{15}{20}= \frac{3}{4}\) and \(\displaystyle \frac{8}{20}= \frac{2}{5}\) so you are still adding the same numbers.
But if you are multiplying, say, 5 by a fraction, you are not just changing the form of a fraction you already have, you are multiplying to produce a new number. I am sure you have learned that \(\displaystyle \frac{a}{b}\times\frac{c}{d}= \frac{ac}{bd}\). Well, \(\displaystyle 5= \frac{5}{1}\) so \(\displaystyle 5(\frac{a}{b})= \frac{5}{1}\frac{a}{b}= \frac{5a}{b}\). Because you did not multiply both numerator and denominator by the same thing you do not have the same fraction now. But you shouldn't have the same fraction- you should have one 5 times as large.
Any calculus book will tel you that the derivative of arcsin(x) is \(\displaystyle \frac{1}{\sqrt{1- x^2}}\) so the derivative of arcsin(u), if u is a function of x, is, by the chain rule, \(\displaystyle \frac{1}{\sqrt{1- u^2}}\frac{du}{dx}\). Here, u(x)= 5x+ 1 so that du/d= 5. The derivative of \(\displaystyle sin^{-1}(5x+10)\) is \(\displaystyle \frac{1}{\sqrt(1- (5x+1)^2}}(5)= \frac{5}{\sqrt{1- 25x^2- 10x+ 1}}= \frac{5}{\sqrt{-25x^2- 10x}}\)
