Math Problem Statement

Y=x^sinx, find dy/dx

Solution

To differentiate y=xsinxy = x^{\sin{x}}, we can use logarithmic differentiation. Here's the process:

  1. Take the natural logarithm of both sides to make the exponent easier to handle: lny=ln(xsinx)\ln{y} = \ln{(x^{\sin{x}})} Using the property of logarithms ln(ab)=blna\ln{(a^b)} = b\ln{a}, this becomes: lny=sinxlnx\ln{y} = \sin{x} \cdot \ln{x}

  2. Differentiate both sides with respect to xx (using implicit differentiation on the left and the product rule on the right): 1ydydx=cosxlnx+sinx1x\frac{1}{y} \cdot \frac{dy}{dx} = \cos{x} \cdot \ln{x} + \sin{x} \cdot \frac{1}{x}

  3. Solve for dydx\frac{dy}{dx} by multiplying both sides by yy: dydx=y(cosxlnx+sinxx)\frac{dy}{dx} = y \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)

  4. Substitute back y=xsinxy = x^{\sin{x}}: dydx=xsinx(cosxlnx+sinxx)\frac{dy}{dx} = x^{\sin{x}} \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)

Thus, the derivative of y=xsinxy = x^{\sin{x}} is: dydx=xsinx(cosxlnx+sinxx)\frac{dy}{dx} = x^{\sin{x}} \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)

Would you like any additional details or further explanation?

Here are 5 related questions:

  1. How would you find the second derivative of y=xsinxy = x^{\sin{x}}?
  2. Can logarithmic differentiation be used for other types of functions? If so, how?
  3. What is the domain of xsinxx^{\sin{x}} for real values of xx?
  4. How does the derivative behave near x=0x = 0?
  5. How can we extend this to functions of the form xf(x)x^{f(x)}?

Tip: When faced with an exponent involving a variable, logarithmic differentiation is a powerful tool to simplify differentiation.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Differentiation
Exponential Functions
Product Rule

Formulas

Logarithmic property: ln(a^b) = b ln(a)
Product rule: d(uv)/dx = u'v + uv'

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Grade 11-12 (Advanced Calculus)