Math Problem Statement

Calculate the partial derivatives ∂f/∂x and ∂f/∂y for the following functions: (a) f(x, y) = tan(x² + y²), (b) f(x, y) = xe^(x²y), (c) f(x, y) = x² ln(y²).

Solution

Let's compute the partial derivatives fx\frac{\partial f}{\partial x} and fy\frac{\partial f}{\partial y} for each function.

(a) f(x,y)=tan(x2+y2)f(x,y) = \tan(x^2 + y^2)

Partial derivative with respect to xx:

fx=sec2(x2+y2)ddx(x2+y2)\frac{\partial f}{\partial x} = \sec^2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2) =sec2(x2+y2)2x= \sec^2(x^2 + y^2) \cdot 2x

Partial derivative with respect to yy:

fy=sec2(x2+y2)ddy(x2+y2)\frac{\partial f}{\partial y} = \sec^2(x^2 + y^2) \cdot \frac{d}{dy}(x^2 + y^2) =sec2(x2+y2)2y= \sec^2(x^2 + y^2) \cdot 2y


(b) f(x,y)=xex2yf(x,y) = x e^{x^2 y}

Partial derivative with respect to xx:

Using the product rule: fx=ex2y+xddx(ex2y)\frac{\partial f}{\partial x} = e^{x^2 y} + x \cdot \frac{d}{dx}(e^{x^2 y}) Using chain rule: ddxex2y=ex2y(2xy)\frac{d}{dx} e^{x^2 y} = e^{x^2 y} \cdot (2xy) fx=ex2y+x(ex2y2xy)\frac{\partial f}{\partial x} = e^{x^2 y} + x (e^{x^2 y} \cdot 2xy) =ex2y(1+2x2y)= e^{x^2 y} (1 + 2x^2 y)

Partial derivative with respect to yy:

fy=xddyex2y\frac{\partial f}{\partial y} = x \cdot \frac{d}{dy} e^{x^2 y} Using chain rule: ddyex2y=ex2y(x2)\frac{d}{dy} e^{x^2 y} = e^{x^2 y} \cdot (x^2) fy=xex2yx2\frac{\partial f}{\partial y} = x e^{x^2 y} x^2 =x3ex2y= x^3 e^{x^2 y}


(c) f(x,y)=x2ln(y2)f(x,y) = x^2 \ln(y^2)

Partial derivative with respect to xx:

fx=2xln(y2)\frac{\partial f}{\partial x} = 2x \ln(y^2)

Partial derivative with respect to yy:

Using chain rule: ddyln(y2)=1y22y=2y\frac{d}{dy} \ln(y^2) = \frac{1}{y^2} \cdot 2y = \frac{2}{y} fy=x22y\frac{\partial f}{\partial y} = x^2 \cdot \frac{2}{y} =2x2y= \frac{2x^2}{y}


Final Answers:

  1. (a)

    • fx=2xsec2(x2+y2)\frac{\partial f}{\partial x} = 2x \sec^2(x^2 + y^2)
    • fy=2ysec2(x2+y2)\frac{\partial f}{\partial y} = 2y \sec^2(x^2 + y^2)
  2. (b)

    • fx=ex2y(1+2x2y)\frac{\partial f}{\partial x} = e^{x^2 y} (1 + 2x^2 y)
    • fy=x3ex2y\frac{\partial f}{\partial y} = x^3 e^{x^2 y}
  3. (c)

    • fx=2xln(y2)\frac{\partial f}{\partial x} = 2x \ln(y^2)
    • fy=2x2y\frac{\partial f}{\partial y} = \frac{2x^2}{y}

Would you like further explanations or step-by-step derivations?


Related Questions:

  1. How do you compute higher-order partial derivatives?
  2. What are the applications of partial derivatives in physics and engineering?
  3. How do you determine if a function is differentiable at a point?
  4. What is the geometric interpretation of partial derivatives?
  5. How do you use the gradient vector in optimization problems?

Tip:

When differentiating functions involving exponentials, logarithms, or trigonometric terms, always check if the chain rule, product rule, or quotient rule applies!

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Multivariable Calculus
Chain Rule
Product Rule

Formulas

∂f/∂x and ∂f/∂y using Chain Rule: ∂f/∂x = df/dg * dg/dx
Product Rule: ∂(uv)/∂x = u'(x)v(x) + u(x)v'(x)

Theorems

Chain Rule for Partial Derivatives
Product Rule for Differentiation

Suitable Grade Level

University-Level (Calculus II or Multivariable Calculus)