Derivative of ex logx:

To find the derivative of ex logx, we can use the product rule of differentiation. The product rule states that if we have two functions, u(x) and v(x), then the derivative of their product, u(x)v(x), is given by:

(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)

In this case, our two functions are ex and logx. Let's differentiate each function separately and then apply the product rule.

Differentiating ex:

The derivative of ex with respect to x is simply ex. This is a well-known property of the exponential function. Therefore, u'(x) = ex.

Differentiating logx:

To differentiate logx, we can use the chain rule. The chain rule states that if we have a function g(x) = f(h(x)), then the derivative of g(x) with respect to x is given by:

d/dx(g(x)) = f'(h(x)) * h'(x)

In this case, our function is logx, which can be written as g(x) = log(x). The inner function is h(x) = x, and the outer function is f(x) = log(x). The derivative of the outer function f(x) = log(x) can be found using the derivative of the natural logarithm, which is 1/x. Therefore, f'(x) = 1/x.

Now, let's apply the chain rule:

d/dx(logx) = f'(h(x)) * h'(x)
= (1/x) * 1
= 1/x

Therefore, v'(x) = 1/x.

Applying the product rule:

Now that we have the derivatives of ex and logx, we can apply the product rule to find the derivative of ex logx.

(d/dx)(ex logx) = u'(x)v(x) + u(x)v'(x)
= ex * logx + ex * (1/x)
= ex(logx + 1/x)

Therefore, the derivative of ex logx is ex(logx + 1/x).

Summary:
- The derivative of ex logx can be found using the product rule.
- The derivative of ex is ex.
- The derivative of logx is 1/x.
- Applying the product rule, we get the derivative of ex logx as ex(logx + 1/x).