Math Problem Statement

determina y'

Solution

To determine yy' from the given equation:

ln(10)log(xy)+cos(xy)=5\ln(10) \cdot \log\left(\frac{x}{y}\right) + \cos(xy) = 5

we will follow these steps:

  1. Implicit Differentiation: Since yy is a function of xx, we differentiate both sides of the equation with respect to xx, treating yy as a function of xx.

  2. Apply Chain Rule: Wherever yy appears, apply the chain rule.

Let’s begin the differentiation.

Step 1: Differentiate both sides with respect to xx:

Differentiating the left side term-by-term:

  • First term: ln(10)log(xy)\ln(10) \cdot \log\left(\frac{x}{y}\right)

    ddx[ln(10)log(xy)]\frac{d}{dx}\left[\ln(10) \cdot \log\left(\frac{x}{y}\right)\right]

    Using the product rule:

    ln(10)ddx[log(xy)]\ln(10) \cdot \frac{d}{dx}\left[\log\left(\frac{x}{y}\right)\right]

    Now, for the derivative of log(xy)\log\left(\frac{x}{y}\right):

    [ \frac{d}{dx}\left[\log\left(\frac{x}{y}\right)\right] = \frac{1}{\frac{x}{y}} \cdot \frac{d}{dx}\left[\frac{x}{y}\right] ]

    Simplifying ddx[xy]\frac{d}{dx}\left[\frac{x}{y}\right]:

    ddx[xy]=yxyy2\frac{d}{dx}\left[\frac{x}{y}\right] = \frac{y - x y'}{y^2}

    Therefore:

    ddx[log(xy)]=yxyx\frac{d}{dx}\left[\log\left(\frac{x}{y}\right)\right] = \frac{y - x y'}{x}

    So, the derivative of the first term is:

    ln(10)yxyx\ln(10) \cdot \frac{y - x y'}{x}

  • Second term: cos(xy)\cos(xy)

    ddx[cos(xy)]\frac{d}{dx}\left[\cos(xy)\right]

    Apply the chain rule:

    sin(xy)ddx(xy)=sin(xy)(y+xy)-\sin(xy) \cdot \frac{d}{dx}(xy) = -\sin(xy) \cdot (y + xy')

The derivative of the left side:

ln(10)yxyxsin(xy)(y+xy)\ln(10) \cdot \frac{y - x y'}{x} - \sin(xy) \cdot (y + xy')

The derivative of the right side 55 is 0.

Step 2: Combine and solve for yy':

Set the derivatives equal:

ln(10)yxyxsin(xy)(y+xy)=0\ln(10) \cdot \frac{y - x y'}{x} - \sin(xy) \cdot (y + xy') = 0

Factor out yy':

[ y'\left[ -\ln(10) + x\sin(xy)\right] = y\left[\sin(xy) - \frac{\ln(10)}{x}\right] ]

Finally, solve for yy':

y=y[sin(xy)ln(10)x]ln(10)+xsin(xy)y' = \frac{y\left[\sin(xy) - \frac{\ln(10)}{x}\right]}{-\ln(10) + x\sin(xy)}

This is the derivative yy' for the given equation.

Do you want more details or have any questions? Here are some related questions to explore:

  1. How would the derivative change if the argument of the logarithm was different?
  2. What would happen if we used a different logarithmic base?
  3. How does implicit differentiation work in cases where both xx and yy are in trigonometric functions?
  4. How can we simplify expressions involving logarithmic and trigonometric functions before differentiation?
  5. How would you interpret the result yy' geometrically?

Tip: Always make sure to apply the chain rule correctly when differentiating composite functions involving products of xx and y(x)y(x).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Logarithmic Functions
Trigonometric Functions

Formulas

Chain rule for implicit differentiation

Theorems

-

Suitable Grade Level

Advanced College