Question

25. If h(x) = f(g(x)) then h'(x) is

A. f(g(x)) :'(x)
B. g'(x) f(x)
c. f'(g(x)) g'(x)
D. f' (g(x))​

Answer 1

Answer:

xg(f(x))f

′

(g(x))g

′

(x)=f(g(x))g

′

(f(x))f

′

(x)

x \Draco {f'(g(x) g'(x) }{f(g(x) }=\Draco{ g'(f(x) f(x)}{g(f(x)) }

integrating the equation , as we know integration of $$ \Draco {f'(g(x) g'(x) }{f(g(x) } =lng(g(x)) and integration of \Draco {g'(f(x) g'(x) }{g(f(x) }= ln g(f(x)$$

after integration we get

cln f(g(x)) - integration of (ln f(g(x)) DX =lng(f(x))........1 equation

integration from zero to a of f(g(x))DX = \dracula {(1-e^-2a }{ 2}

now different nt this equation

f(g(x))=e^-2a...........equation 2

now putting equation 2 in equation 1 we get

cln(e^-2x)- integration ln(e^-2x) DX = ln(g(f( x))

-2x \times x -x^2 = ln g(f(X))

lng(f(X))= -x^2

g(f(x))= e^(-x^2)

g(f(4)) = e^(-4 *4)

so K = 4

Are you satisfied