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Derivatives & Second-Order Derivatives · Question 63 of 96

Question 63

EasyOriginal question

If y=sin⁡2xy = \sin 2x, what is d2ydx2\frac{d^2y}{dx^2}?

Options for question 63

Show the answer and explanation
CorrectNot quiteAnswer D

Correct answer: D

Explanation

  1. First derivative: dydx=2cos⁡2x\frac{dy}{dx} = 2 \cos 2x.

  2. Differentiate again: d2ydx2=−4sin⁡2x\frac{d^2y}{dx^2} = - 4 \sin 2x.

Worth noting

2cos⁡2x2 \cos 2x stops at the first derivative; d2ydx2\frac{d^2y}{dx^2} differentiates it once more.

Concept
Second derivative of a sine or cosine
Typical time
60 s
Your time
—
The rule this testsSecond-order derivatives
Practise this chapter →

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

Practise this chapter →

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.