Math Problem Statement

ln(x^2 + 5y) = sin(xy)

Solution

The equation given for implicit differentiation is:

ln(x2+5y)=sin(xy)\ln(x^2 + 5y) = \sin(xy)

To find the derivative dydx\frac{dy}{dx}, we will differentiate both sides of the equation with respect to xx, applying the chain rule where necessary.

Step-by-step solution:

Left-hand side:

ddx(ln(x2+5y))\frac{d}{dx} \left( \ln(x^2 + 5y) \right) Using the chain rule, this becomes: 1x2+5y(2x+5dydx)\frac{1}{x^2 + 5y} \cdot \left( 2x + 5 \frac{dy}{dx} \right) Here, 2x2x is the derivative of x2x^2, and 5dydx5 \frac{dy}{dx} is the derivative of 5y5y with respect to xx, where we applied the chain rule to differentiate yy implicitly.

Right-hand side:

ddx(sin(xy))\frac{d}{dx} \left( \sin(xy) \right) Using the product and chain rules, we get: cos(xy)(y+xdydx)\cos(xy) \cdot \left( y + x \frac{dy}{dx} \right) Here, yy is the derivative of xyxy with respect to xx, and xdydxx \frac{dy}{dx} is the implicit differentiation of yy.

Combining both sides:

Now, we have the equation: 1x2+5y(2x+5dydx)=cos(xy)(y+xdydx)\frac{1}{x^2 + 5y} \cdot \left( 2x + 5 \frac{dy}{dx} \right) = \cos(xy) \cdot \left( y + x \frac{dy}{dx} \right)

Solving for dydx\frac{dy}{dx}:

To isolate dydx\frac{dy}{dx}, expand both sides and group all terms containing dydx\frac{dy}{dx} on one side.

Would you like to continue with the full solution, or any specific part of this step detailed further?

Here are some questions related to this process:

  1. Why do we apply the chain rule when differentiating ln(x2+5y)\ln(x^2 + 5y)?
  2. What is the role of implicit differentiation in handling yy in the equation?
  3. How does the product rule apply to the term sin(xy)\sin(xy)?
  4. Could you solve for dydx\frac{dy}{dx} explicitly by expanding both sides of the equation?
  5. What would change in the process if the equation involved cos(x2+y2)\cos(x^2 + y^2)?

Tip: When performing implicit differentiation, always treat yy as a function of xx and apply the chain rule when differentiating terms involving yy.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Product Rule
Logarithmic Differentiation
Trigonometric Differentiation

Formulas

Chain Rule: d(uv)/dx = u'(v) + u(v')
Logarithmic Differentiation: d/dx(ln(u)) = (1/u) * du/dx
Trigonometric Derivative: d/dx(sin(u)) = cos(u) * du/dx

Theorems

Implicit Differentiation
Chain Rule
Product Rule

Suitable Grade Level

Grades 10-12