#calculus proof

[roomfullofpigeons]

Differentiate f(x) = g(x)h(x) [Product rule proof]

f(x) = g(x)h(x)  [I’m defining the function “f(x)” as the product of two other functions, g(x) and h(x)]

ln(f(x)) = ln[g(x)h(x)] = ln[g(x)] +ln[h(x)] [I took the natural logarithm of both sides, so that I could split up my product into a sum of natural logarithms.]

f’(x)/f(x) = g’(x)/g(x) + h’(x)/h(x) = [g’(x)h(x) + g(x)h’(x)]/g(x)h(x)  [I’ve differentiated both sides, then written my side with g(x) and h(x) as a single fraction]

f’(x)/f(x) = [g’(x)h(x) + g(x)h’(x)]/f(x) [Substituted in f(x)=g(x)h(x)]

f’(x) =  g’(x)h(x) + g(x)h’(x)  [Multiply both sides by f(x). Now we have a formula for the differential of g(x)h(x)! This is called the product rule!]

[roomfullofpigeons]

Integration by parts!

Assuming product rule for differentiation is true (here’s a link to a nice little proof for that), we can get the rule for integration by parts!

If f(x) = g(x)h(x), then f’(x) =  g’(x)h(x) + g(x)h’(x)  [This is just the product rule] 

f(x) +c =  ∫[g’(x)h(x) + g(x)h’(x)].dx  [I’ve integrated both sides with respect to x indefinitely.]

f(x) +c =  ∫g’(x)h(x).dx + ∫g(x)h’(x).dx [I’ve split my integral into a sum of two integrals. The area under two curves added together is the same as adding the areas under each curve. Also, this is how you would integrate a polynomial.]

g(x)h(x) + c = ∫g’(x)h(x).dx + ∫g(x)h’(x).dx  [I’ve substituted f(x) = (g(x)h(x)]

g(x)h(x) + c - ∫g’(x)h(x).dx  =  ∫g(x)h’(x).dx [subtract  ∫g’(x)h(x).dx from both sides]

∫g(x)h’(x).dx = g(x)h(x) + c - ∫g’(x)h(x).dx [Flip it so that the equation reads in the direction we want it to. It’s ot needed; it just makes things look a little bit nicer]

∫g(x)h’(x).dx = g(x)h(x) - ∫g’(x)h(x).dx [We don’t need the “+c” because there is an integral on the right side of the equation which will add a constant anyway.]

∫g(x)h’(x).dx = g(x)h(x) - ∫g’(x)h(x).dx  This is the formula for integration by parts.