Question

$\frac{d\sin x^2}{dx}$

• A. $2x\cos x^2$
• B. $4x\cos x^2$
• C. $2x\sin x^2$
• D. $-2x\sin x^2$

Answer 1

Answer: (A)

$y=\sin x^2$
$\frac{dy}{dx}=\underset{\delta x\to 0}{lim}\frac{\sin {(x+\delta x)}^2-\sin x^2}{\delta x}$
$=\underset{\delta x\to 0}{lim}\frac{\sin (x^2+2x\delta x^2+(\delta x{)}^2)-\sin x^2}{\delta x}$
$=\underset{\delta x\to 0}{lim}\frac{2\cos (x^2+x\delta x+\frac{1}{2}\delta x^2)\sin (x\delta x+\frac{1}{2}\delta x^2)}{\delta x}$
$=2\cos x^2\underset{\delta x\to 0}{lim}\frac{\sin (x\delta x+\frac{1}{2}\delta x^2)}{x\delta x+\frac{1}{2}\delta x^2}.\frac{x\delta x+\frac{1}{2}\delta x^2}{\delta x}$
$=2\cos x^2.1.x=2x\cos x^2$