more derivatives

another thing I did last night was solve for the derivative function of a semicircle curve f(x)=sqrt(1-x^2)

I thought hey I'll decompose this into two functions
f(x)=g(h(x))
where g(x) = sqrt(x)
and h(x) = 1-x^2

g'(x) = sqrt(x) should be first.

I know that d/dx f^-1(x) = 1/f'(f^-1(x))
and obviously that d/dx x^2 = 2x
and almost too obvious to even have to say that sqrt(x) is the inverse function of x^2.
so d/dx sqrt(x) = 1/(2*sqrt(x))

now, h'(x) = d/dx * 1-x^2 = d/dx * -x^2 = -2x

so, d/dx * g(h(x)) = g'(h(x))*h'(x)
=-2x/(2*sqrt(1-x^2))
= -x/sqrt(1-x^2)

ding ding ding that's the formula for the slope of a circle in terms of x.
if you wanted in terms of angle just use tangent heheh...

math is fun

I was yesterday doing some basic differential calculus and so here I shall share it.
f(x) = 1/x
f'(x) = d/dx * 1/x
= [1/(x+h) - 1/x]/h
= [x/(x(x+h)) - (x+h)/(x(x+h))]/h
= [(x-(x+h)) / (x(x+h))] /h
= -h / (x(x+h)) / h
= -1/(x^2+hx)

lim (-1/(x^2+hx)) = -1/(x^2+0x) = -1/x^2
h->0

f'(x) = d/dx * 1/x = -1/x^2

yay.