JAMB - Mathematics (1998 - No. 28)
Differentiate \(\frac{x}{cosx}\) with respect to x
1 + x sec x tan x
1 + sec2 x
cos x + x tan x
x sec x tan x + secx
Explication
let y = \(\frac{x}{cosx}\) = x sec x ( since \(\frac{1}{cosx}\) = sec x )
let u = x, v = sec x
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
\(\frac{dy}{dx}\) = x [secx tanx] + secx
Therefore, \(\frac{dy}{dx}\)\(\frac{x}{cosx}\) = x secx tanx + secx
let u = x, v = sec x
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
\(\frac{dy}{dx}\) = x [secx tanx] + secx
Therefore, \(\frac{dy}{dx}\)\(\frac{x}{cosx}\) = x secx tanx + secx