Question

If f '(1)=2, $g'(\\\sqrt{2})$ = 4 then the derivative of

f (tan x) with respect to g(sec x) at x = p/4 is

• A. 1
• B. $\sqrt{2}$
• C. $\frac{\sqrt{2}}{2}$
• D. 2

Answer 1

Answer: (C)

$\frac{\sqrt{2}}{2}$

Answer 2

y = f(tan x) z = g(sec x)

$\frac{dy}{dx}$= f '(tanx).sec²x

$\frac{dz}{dx}$= g'(sec x) secx tanx

put x = $\frac{\pi}{4}$

$\frac{dy}{dx}$ = f ' (1).2

$\frac{dz}{dx}$=$g'(\sqrt{2}).\sqrt{2}$

$\frac{dy}{dz}$=$\frac{2f'(1)}{g'(\sqrt{2}).\sqrt{2}}$ = $\frac{2 \times 2}{4\sqrt{2}}$=$\frac{1}{\sqrt{2}}$