Answer to Question #118866 in Calculus for Nimra

Question

f³f¹(x) = √secx.x³

Answer 1

$$f(x)=\sqrt{\sec x}\cdot x^3$$

$${df\over dx}={x^3\over 2\sqrt{\sec x}}(-{-\sin x\over \cos^2x})+3x^2\sqrt{\sec x}=$$

$$=x^2{x\tan x+6\over 2\cos x\sqrt{\sec x}}=x^2{x\tan x+6\over 2\sqrt{\cos x}}=$$

$$=x^2\sqrt{\sec x}\cdot{x\tan x+6\over 2}$$