Find the gradient of the curve y=20/x²+1 at the point(3,2) Step by step

Mathematics

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Answer

−65-\frac{6}{5}

Graph of y = 20/x² + 1

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Key points of y = 20/x² + 1
Pointxy

Step-by-step solution

4 steps

  1. Rewrite the function for easier differentiation.

    The given function is y=20x2+1y = \frac{20}{x^2+1}.
    This can be written as y=20(x2+1)−1y = 20(x^2+1)^{-1}.

  2. Differentiate the function with respect to xx.

    We use the chain rule: dydx=ddx[20(x2+1)−1]\frac{dy}{dx} = \frac{d}{dx} [20(x^2+1)^{-1}].

    dydx=20⋅(−1)(x2+1)−2⋅ddx(x2+1)\frac{dy}{dx} = 20 \cdot (-1)(x^2+1)^{-2} \cdot \frac{d}{dx}(x^2+1) dydx=−20(x2+1)−2⋅(2x)\frac{dy}{dx} = -20(x^2+1)^{-2} \cdot (2x) dydx=−40x(x2+1)2\frac{dy}{dx} = -\frac{40x}{(x^2+1)^2}

  3. Substitute the xx-coordinate of the given point into the derivative.

    The given point is (3,2)(3,2), so x=3x=3.

    dydx∣x=3=−40(3)((3)2+1)2\frac{dy}{dx}\Big|_{x=3} = -\frac{40(3)}{((3)^2+1)^2} dydx∣x=3=−120(9+1)2\frac{dy}{dx}\Big|_{x=3} = -\frac{120}{(9+1)^2} dydx∣x=3=−120(10)2\frac{dy}{dx}\Big|_{x=3} = -\frac{120}{(10)^2} dydx∣x=3=−120100\frac{dy}{dx}\Big|_{x=3} = -\frac{120}{100}

  4. Simplify the result.

    dydx∣x=3=−1210=−65\frac{dy}{dx}\Big|_{x=3} = -\frac{12}{10} = -\frac{6}{5}

    The gradient of the curve at the point (3,2)(3,2) is −65\boxed{-\frac{6}{5}}.

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