# Derivative of the exponential term

The equation for blackbody radiation (the Planck equation) is:

d(uv) = u dv + v du

where:

v = [exp(c2/( T) -1)]-1

We need to take the derivative of this term. There are two possibilities:

1. as an expression raised to a power: v = Zt where dv = t Zt - 1 dZ or
2. as a fraction : v = 1/Z where dv = [Z  × d(1) - 1  × dZ]/Z2
In either case, we should obtain the same answer. Let us try.

Case 1 here the power, t, equals -1. Immediately we have dv = -1   [exp(c2/( T) -1)]-2 dZ. Now we must find the derivative of the exponential term. The devivative of an exponential is simply the exponential term itself multiplied by the derivative of the exponent. In this case, since we are taking the derivative with respect to wavelength, , the derivative of the exponent is -[c2/ 2 T]exp(c2/( T). Combining terms, we get:

dv = [c2/( 2 T)] [exp(c2 / T)/(exp(c2 / T) - 1)2].

Case 2: the derivative of 1, a constant, is 0, so the first term in the numerator drops out. For the second term, we must again know what the derivative of exp(c2/( T) - 1) is. Using our result above, combining these terms (and noticing that they will be multiplied by -1), we again get the expression below.

dv = [c2/( 2 T)] [exp(c2 / T)/(exp(c2 / T) - 1)2].

Not surprisingly, they are identical.

Note: The notation exp(x) is another way of writing ex. In this case I used this form because of the complexity of the term in the exponent, the difficulty of properly creating it in HTML (the language used to write these pages), and the relatively low resolution of computer screens.