Differentiating y = x^sinx(logx)^x

To find dy/dx, we need to differentiate the given function y = x^sinx(logx)^x with respect to x. Let's break down the process step by step.

Step 1: Rewrite the function using exponential and logarithmic properties
We can rewrite the function as y = e^(sinx * ln(x^x)).

Step 2: Apply the chain rule
Since we have a composition of functions, we need to apply the chain rule to differentiate y with respect to x.

The chain rule states that if we have a composite function y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

Step 3: Differentiate the outer function
In our case, the outer function is e^u, where u = sinx * ln(x^x).

To differentiate e^u with respect to u, we simply multiply by the derivative of u.

Therefore, dy/du = e^u * d(u)/dx.

Step 4: Differentiate the inner function
Now, let's determine the derivative of u = sinx * ln(x^x) with respect to x.

To differentiate sinx, we get d(sinx)/dx = cosx.

To differentiate ln(x^x), we apply the logarithmic differentiation rule, which states that d(ln(u))/dx = (1/u) * du/dx.

In our case, u = x^x, so we have d(ln(x^x))/dx = (1/(x^x)) * d(x^x)/dx.

Using the power rule, d(x^x)/dx = x^x * (1 + ln(x)).

Therefore, d(ln(x^x))/dx = (1/(x^x)) * x^x * (1 + ln(x)) = 1 + ln(x).

Finally, d(u)/dx = d(sinx)/dx * ln(x^x) + sinx * d(ln(x^x))/dx = cosx * ln(x^x) + sinx * (1 + ln(x)).

Step 5: Calculate dy/dx
Now that we have dy/du and du/dx, we can calculate dy/dx.

dy/du = e^u * (cosx * ln(x^x) + sinx * (1 + ln(x))).

dy/dx = dy/du * du/dx = e^u * (cosx * ln(x^x) + sinx * (1 + ln(x))) * (cosx * ln(x^x) + sinx * (1 + ln(x))).

Therefore, the derivative of y = x^sinx(logx)^x with respect to x is dy/dx = e^u * (cosx * ln(x^x) + sinx * (1 + ln(x))) * (cosx * ln(x^x) + sinx * (1 + ln(x))).