Final answer:
The derivative of [sin(x)]^x is found by first applying the natural logarithm to both sides, using implicit differentiation, and then solving for dy/dx, which results in dy/dx = [sin(x)]^x × (ln(sin(x)) + x × cot(x)).
Step-by-step explanation:
To find the derivative of [sin(x)]^x, we need to apply the chain rule and implicit differentiation since the function is quite complex. First, we will consider the function as y = [sin(x)]^x and take the logarithm of both sides to simplify the differentiation process. Applying the natural logarithm, we get ln(y) = x ln(sin(x)). Differentiating implicitly with respect to x, we obtain:
1/y × dy/dx = ln(sin(x)) + x/cos(x).
Now, we solve for dy/dx by multiplying both sides by y which gives us:
dy/dx = y(ln(sin(x)) + x/cos(x))
Substituting back the original function for y, we have:
dy/dx = [sin(x)]^x × (ln(sin(x)) + x × cot(x))
This derivative showcases both logarithmic differentiation and the chain rule. However, this question involves terms with partial derivatives and trigonometric identities used in physics, which are not directly related to finding the derivative of [sin(x)]^x. So those references are not required to answer this question about derivatives in the context of mathematics.