How To Derive The Chain Rule Formula For Differentiation Of Logarithms With Base ‘A’

d/dx[loga(x)]

1/xlna

We can use the chain rule of differentiation to find the derivative of loga(x) with respect to x.

Let u = loga(x). Then,

du/dx = d/dx [loga(x)]

To apply the chain rule, we need to find du/dx in terms of x.

We know that loga(x) = ln(x)/ln(a) (where ln denotes the natural logarithm). Therefore,

u = ln(x)/ln(a)

Using the quotient rule of differentiation, we can find du/dx:

du/dx = (1/ln(a)) * [d/dx(ln(x))] – (ln(x)/ln^2(a)) * [d/dx(ln(a))]

Since d/dx(ln(a)) = 0 (because a is a constant), the second term in the equation above becomes zero:

du/dx = (1/ln(a)) * [d/dx(ln(x))]

Now, we need to find d/dx(ln(x)). Again, using the chain rule, let v = ln(x):

dv/dx = d/dx[ln(x)]

Using the inverse function rule of differentiation, we know that d/dx[ln(x)] = 1/x. Therefore,

dv/dx = 1/x

Substituting this back into the expression for du/dx, we get:

du/dx = (1/ln(a)) * [dv/dx]

Substituting back v = ln(x), we get:

du/dx = (1/ln(a)) * (1/x)

Therefore,

d/dx[loga(x)] = du/dx = (1/ln(a)) * (1/x)

Simplifying the expression, we get:

d/dx[loga(x)] = 1/(x*ln(a))

More Answers: