Question

What is the slope of the tangent to the curve y = 2x/(x^2 + 1) at (0, 0)?

(a) 0

(b) 1

(c) 2

(d) 3

I had been asked this question in quiz.

I'd like to ask this question from Application of Derivative topic in chapter Application of Derivatives of Mathematics – Class 12

Answer 1

Correct answer is (c) 2

The explanation is: We have y = 2x/(x^2 + 1)

Differentiating y with respect to x, we get

dy/dx = d/dx(2x/(x^2 + 1))

= 2 * [(x^2 + 1)*1 – x * 2x]/(x^2 + 1)^2

= 2 * [1 – x^2]/(x^2 + 1)^2

Thus, the slope of tangent to the curve at (0, 0) is,

[dy/dx](0, 0) = 2 * [1 – 0]/(0 + 1)^2

Thus [dy/dx](0, 0) = 2.