Calculus Gymnasiematematik(high school math) matematik 3c matematik 4

Differentiating the natural logarithm, products and quotients

In order to be able to deduce the derivative of the natural logarithm we resort to using implicit differentiation.

Let x= ey(x)

Differentiating both sides gives

dx/dx = d ey(x)/dx

1=ey(x) dy(x)/dx

Solving for dy(x)/dx one obtains

dy(x)/dx = 1/ey(x) = 1/x .

The product rule is given by

d f(x) g(x)/dx = df(x)/dxg(x) + f(x) dg(x)/dx

A beautiful proof for this theorem is given by G.W. Leibniz:

(u + du) *(v+dv) = u*v + u*dv + v* du + du*dv.

The last term is the product of two infinitesimals and can therefore be neglected. The differential of the product of the two functions u(x)*v(x) is thus equal to

u(x)*dv + v(x)*du.


To read more about this formula click here.

The rule for differentiation of the quotient of  two functions can be deduced from the product rule and is given by

df(x)/g(x) =( df(x)/dx g(x) – f(x) dg(x))/dx)/(g(x))2.

Av mattelararen

Licentiate of Philosophy in atomic Physics
Master of Science in Physics


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