Slope of a curve

Calculus Level 2

Find the slope of the curve y = x x 2 + 4 y=\dfrac{x}{x^2+4} at point ( 2 , 1 4 ) \left(2,\dfrac{1}{4}\right) .

1 4 0 -2

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1 solution

The derivative of a function is identical with the slope of the graph of the function. By the quotient rule, we have

y = x x 2 + 4 y=\dfrac{x}{x^2+4}

d y d x = ( x 2 + 4 ) ( 1 ) − x ( 2 x ) ( x 2 + 4 ) 2 = x 2 + 4 − 2 x 2 ( x 2 + 4 ) 2 = 4 − x 2 ( x 2 + 4 ) 2 \dfrac{dy}{dx}=\dfrac{(x^2+4)(1)-x(2x)}{(x^2+4)^2}=\dfrac{x^2+4-2x^2}{(x^2+4)^2}=\dfrac{4-x^2}{(x^2+4)^2}

At x = 2 x=2

d y d x = 4 − 2 2 ( 2 2 + 4 ) 2 = 0 \dfrac{dy}{dx}=\dfrac{4-2^2}{(2^2+4)^2}=\boxed{0}

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