Question

If u = f(x²), v = g(x³) ; f ′(x) = sin x, g ′(x) = cos x, then $\frac{du}{dv}$ =

• A. $\frac{2\sin(x^{2})}{3x(\cos x^{3})}$
• B. $\frac{2}{3x}$ tan x
• C. $\frac{2}{3x}\frac{\sin^{2}x}{\cos^{3}x}$
• D. None

Answer 1

Answer: (A)

$\frac{2\sin(x^{2})}{3x(\cos x^{3})}$

Answer 2

$\frac{du}{dv} = \frac{du/dx}{dv/dx}$ =$\frac{f'(x^{2}).2x}{9'(x^{3}).3x^{2}} = \frac{\sin x^{2}.2x}{\cos x^{3}.3x^{2}}$= $\frac{2\sin x^{2}}{3x\cos x^{3}}$