how do you differentiate log base5(√(7x/3x+2)^ln5)?

Answer 1

First convert the base 5 log to natural; this simply involves a constant

ln(x) = log bas 5 (x) / ln(5), so log base 5 (x) = ln(5)*ln(x):

so now differentiate ln(5)*ln[√(7x/3+2)^ln(5)]

Note that √(7x/3+2)^ln(5) = (7x/3+2)^(ln(5)/2)

The rest is a series of chain-rule applications:

ln(5) * [7x/3+2)^-ln(5)/2] * d/dx (7x/3+2)^ln(5)/2

d/dx [(7x/3+2)]^ln(5)/2 = ln(5)/2 * [(7x/3+2)]^(ln(5)/2-1) * d/dx (7x/3+2)

d/dx (7x/3+2) = 7/3

combine the parts

ln(5) * [(7x/3+2)^-ln(5)/2] * ln(5)/2 * [(7x/3+2)]^(ln(5)/2-1) * 7/3

The terms containing (7x/3+2) can be combined as can the constants

[7*(ln(5)^2 / 6] * (7x/3+2)^(-1)

[7*(ln(5)^2 / 6] / (7x/3+2)