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Let 
y=f(x) be a differentiable function such that 
(dy)/(dx)=(x)/(y) and 
f(8)=2. What is the approximation of 
f(8.1) using the line tangent to the graph of 
f at 
x=8 ?

Let y=f(x) y=f(x) be a differentiable function such that dydx=xy \frac{d y}{d x}=\frac{x}{y} and f(8)=2 f(8)=2 . What is the approximation of f(8.1) f(8.1) using the line tangent to the graph of f f at x=8 x=8 ?

Full solution

Q. Let y=f(x) y=f(x) be a differentiable function such that dydx=xy \frac{d y}{d x}=\frac{x}{y} and f(8)=2 f(8)=2 . What is the approximation of f(8.1) f(8.1) using the line tangent to the graph of f f at x=8 x=8 ?
  1. Identify derivative & point: Identify the derivative and the point of interest.\newlineGiven that dydx=xy\frac{dy}{dx} = \frac{x}{y} and f(8)=2f(8) = 2, we need to find the derivative at x=8x = 8.
  2. Calculate derivative at x=8x=8: Calculate the derivative at x=8x = 8.dydx\frac{dy}{dx} at x=8x = 8 is 82=4\frac{8}{2} = 4.
  3. Use point-slope form: Use the point-slope form of the equation of a line to find the tangent line.\newlineThe slope of the tangent line at x=8x = 8 is 44, and the point on the curve is (8,2)(8, 2).\newlineEquation of the tangent line: y2=4(x8)y - 2 = 4(x - 8).
  4. Simplify tangent line equation: Simplify the equation of the tangent line. \newliney2=4x32y - 2 = 4x - 32,\newliney=4x30y = 4x - 30.
  5. Approximate f(8.1)f(8.1): Use the tangent line to approximate f(8.1)f(8.1).\newlineSubstitute x=8.1x = 8.1 into the tangent line equation:\newliney=4(8.1)30y = 4(8.1) - 30,\newliney=32.430y = 32.4 - 30,\newliney=2.4y = 2.4.

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