## d/dx log a(x)

### To differentiate the logarithm function with respect to x, specifically log base a of x (logₐ(x)), we can use the chain rule

To differentiate the logarithm function with respect to x, specifically log base a of x (logₐ(x)), we can use the chain rule.

Let’s denote the logarithmic function as y = logₐ(x), where a is the base.

Using the definition of logarithms, we can rewrite the equation as a^y = x. Now, we will differentiate both sides of this equation with respect to x.

On the left-hand side, we have a^y. To differentiate this, we need to express it in terms of x. Since a^y = x, we can take the logarithm of both sides using base a, giving us logₐ(a^y) = logₐ(x). As logₐ(a^y) simplifies to y (since logₐ(a^y) = y), we have y = logₐ(x).

Now, differentiating both sides of y = logₐ(x) with respect to x using the chain rule:

d/dx(y) = d/dx(logₐ(x))

To find d/dx(y), we need to find dy/dx.

Using the natural logarithm (ln) to represent the logarithm with base e, we have y = ln(x) / ln(a), where ln(a) is a constant.

To differentiate y = ln(x) / ln(a), we’ll use the quotient rule:

dy/dx = (ln(a) d/dx(ln(x)) – ln(x) d/dx(ln(a))) / (ln(a))^2

Now, let’s find the derivatives in the numerator:

d/dx(ln(x)) = 1/x (derivative of ln(x))

d/dx(ln(a)) = 0 (derivative of a constant)

Substituting these derivatives back into the equation:

dy/dx = (ln(a) * (1/x) – ln(x) * 0) / (ln(a))^2

dy/dx = ln(a) / (x * ln(a))^2

dy/dx = ln(a) / (ln(a))^2 * (1/x)^2

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

Therefore, d/dx(logₐ(x)) = 1 / (x * ln(a)).

In simpler terms, the derivative of logₐ(x) with respect to x is 1 divided by the product of x and the natural logarithm of the base a.

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