If you want to use implicit differentiation rather than the chain rule we have: $$x^2+y^2=1$$ $$(x+dx)^2+(y+dy)^2=1$$ $$x^2+2x.dx+dx^2+y^2+2y.dy+dy^2=1$$ Neglecting nilsquare terms and substituting for $y^2$ and $y$: $$x^2+2x.dx+(1-x^2)+2(\sqrt{1-x²})dy=1$$ $$\frac{dy}{dx}=\frac{-x}{\sqrt{1-x²}}$...

A difficult question. Since it was me who pulled the trigger, let me give a “confession”. I have seen mainly three different kinds of "bad" questions in MSE so far: Spams; please-do-my-homework-I-don't-know-anything-about-it type questions; the question is mathematically ill-defined but might...

$$f'(x)=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$ $$[2]^{2'}=\frac{(2+h)^2-2^2}{h}=4+h$$ $$(\frac{1}{[2]})'=\frac{\frac{1}{2+h}-\frac{1}{2}}{h}=\frac{\frac{-h}{4+2h}}{h}=\frac{-1}{4+2h}$$ NB First answer given.

How can I get $$\sum_{k=0}^n\binom{n}{k}(pe^t)^kq^{n-k}=(pe^t+q)^n?$$ I'm having a real problem, trying to figure out this equation. Can you, help me with that ?