Calculus (derivatives)

Question

I am having trouble with these questions

Let F (x) = f (x f (x2)). Suppose that f (4) = 6, f '(4) = 2, and f '(12) = 4. Find F '(2).	Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 6. If the second approximation is found to be x2 = −5, and the tangent line to f (x) at x = 6 passes through the point (17, 3), find f (6).	Use Newton's method with initial approximation x1 = − 6 to find x2, the second approximation to the root of the equation x3 = 7x − 9.

Answer 1

Hi,

F (x) = f (x f (x2)). Suppose that f (4) = 6, f '(4) = 2, and f '(12) = 4. Find F '(2)

First, you need to know:

#1 f'(u) = u'(x) .f'( u) that u is a function of x

#2 (f.g)' = f'g + g'f rule of product

F (x) = f (x f (x2)) in this case x f (x2) is u that I told you at #1

now find u'(x) = ( x f (x2) ) ' = 1 * f(x²) + x * (f(x²))' follow product rule

one more time follow #1 for ( f(x²) )' = 2x * f'(x²) must be replace!

now taking derivative of both sides F (x) = f (x f (x2))

F'(x) = [f (x f (x2))]' = [1 * f(x²) + x * (f(x²))' ] *f' (x f (x2))

= [ 1 * f(x²) + x * ( 2x * f'(x²) )] * f' (x f (x2))

F'(2) = [ 1 *f(4) + 2(4*f'(4))] * f'(2*f(4)) be careful 2f(4) = 12

F'(2) = [ 1*6 + 2*4*2] * f'(12) you know f'(12) =4

F'(2) = [6+16] * 4 = 88

I hope it is useful,

Minoo