Question

Given = 3 + 7, = + 1, = 2 − 3, find f(2) + h(1) and f(0) + g(0) - h(0)​

Answer 1

Answer:

$hi$

Step-by-step explanation:

y=fgh

⇒

dx

dy

=f

′

gh+fg

′

h+fgh

′

=

2

1

(2f

′

gh+2fg

′

h+2fgh

′

)

=

2

1

(h(f

′

g+g

′

f)+g(f

′

h+fh

′

)+f(g

′

h+gh

′

))

=

2

1

(h.(fg)

′

+g.(fh)

′

+f.(gh)

′

)

∴(fgh)

′

(0)=

2

1

(h(0)(fg)

′

(0)+g(0)(fh)

′

(0)+f(0)(gh)

′

(0))

=

2

1

(3×6+2×5+1×4)=

2

1