Math Problem Statement

implicit differentiation of: x^8y^9-y=x

Solution

To perform implicit differentiation on the equation:

x8y9y=xx^8 y^9 - y = x

we will differentiate both sides with respect to xx, treating yy as a function of xx (i.e., y=y(x)y = y(x)).

Step-by-Step Solution:

  1. Differentiate the left-hand side:

ddx(x8y9y)\frac{d}{dx}\left(x^8 y^9 - y\right)

  • For the term x8y9x^8 y^9, we apply the product rule: ddx(x8y9)=ddx(x8)y9+x8ddx(y9)\frac{d}{dx}(x^8 y^9) = \frac{d}{dx}(x^8) \cdot y^9 + x^8 \cdot \frac{d}{dx}(y^9) =8x7y9+x89y8dydx= 8x^7 y^9 + x^8 \cdot 9y^8 \frac{dy}{dx}

  • For the term y-y, we differentiate with respect to xx, treating yy as a function of xx: ddx(y)=dydx\frac{d}{dx}(-y) = -\frac{dy}{dx}

Thus, the derivative of the left-hand side is: 8x7y9+9x8y8dydxdydx8x^7 y^9 + 9x^8 y^8 \frac{dy}{dx} - \frac{dy}{dx}

  1. Differentiate the right-hand side:

ddx(x)=1\frac{d}{dx}(x) = 1

Now, we have the equation: 8x7y9+9x8y8dydxdydx=18x^7 y^9 + 9x^8 y^8 \frac{dy}{dx} - \frac{dy}{dx} = 1

  1. Solve for dydx\frac{dy}{dx}:

Group all the terms involving dydx\frac{dy}{dx}: 9x8y8dydxdydx=18x7y99x^8 y^8 \frac{dy}{dx} - \frac{dy}{dx} = 1 - 8x^7 y^9

Factor out dydx\frac{dy}{dx} on the left-hand side: dydx(9x8y81)=18x7y9\frac{dy}{dx}(9x^8 y^8 - 1) = 1 - 8x^7 y^9

Finally, solve for dydx\frac{dy}{dx}: dydx=18x7y99x8y81\frac{dy}{dx} = \frac{1 - 8x^7 y^9}{9x^8 y^8 - 1}

Thus, the derivative is: dydx=18x7y99x8y81\boxed{\frac{dy}{dx} = \frac{1 - 8x^7 y^9}{9x^8 y^8 - 1}}

Would you like further details or have any questions about the steps?


Here are 5 related questions to expand your understanding:

  1. How would you apply implicit differentiation to x3y2+sin(y)=ln(x)x^3y^2 + \sin(y) = \ln(x)?
  2. What is the significance of using the product rule when implicitly differentiating products of xx and yy?
  3. Can you find the second derivative of yy with respect to xx for the original equation?
  4. What happens to the solution if x=1x = 1 and y=1y = 1 in the given equation?
  5. How does implicit differentiation differ from explicit differentiation?

Tip: When implicitly differentiating, always remember to apply the chain rule to the terms involving yy, since yy is treated as a function of xx.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Product Rule
Chain Rule

Formulas

Product Rule: d(uv)/dx = u'(v) + u(v')
Chain Rule: d(f(g(x)))/dx = f'(g(x)) * g'(x)

Theorems

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Suitable Grade Level

Grades 11-12 or Early College