Question

Solve the following : 

The values of f(x), g(x), f'(x) and g'(x) are given in the following table :

x	f(x)	g(x)	f'(x)	fg'(x)
– 1	3	2	– 3	4
2	2	– 1	– 5	– 4

Match the following :

A Group – Function	B Group – Derivative
(A)`"d"/"dx"[f(g(x))]"at" x = -1`	1.  – 16
(B)`"d"/"dx"[g(f(x) - 1)]"at" x = -1`	2.     20
(C)`"d"/"dx"[f(f(x) - 3)]"at" x = 2`	3.  – 20
(D)`"d"/"dx"[g(g(x))]"at"x = 2`	5.     12

Answer 1

(A) `"d"/"dx"[f(g(x))]` 
= `f'(g(x))."d"/"dx"(g(x))`
= f'(g(x)) xg'(x)

∴ `"d"/"dx"[f(g(x))]` at x = – 1
= f'(g(– 1)) x g'(– 1)
= f'(2) x g'(– 1)   ...[∵ g(x) = 2, when x = – 1]
= – 5 x 4
= – 20

(B) `"d"/"dx"[g(f(x) - 1)]`

= `g'(f(x) - 1)."d"/"dx"[f(x) - 1]`
= g'(f(x) – 1) x [f'(x) –  0]

∴ `"d"/"dx"[gf(x) - 1]` at x = – 1
= g'(f(– 1)– 1) xx f'( –1)
= g'(3 – 1) x f'(– 1)  ...[∵ f(x) 33, when x = – 1]
= g'(2) x f'(– 1)
= (– 4)(– 3)
= 12

(C) `"d"/"dx"[f(f(x) - 3)]`

= `f'(f(x) - 3)."d"/"dx"[f(x) - 3]`
= f'(f(x) – 3) x [f'(x) – 0]

∴ `"d"/"dx"[f(f(x) - 3)]` at = 2
= f"(f(2) – 3) x f'(2)
= f'(2 – 3) x f'(2)    ...[∵  f(x) = 2, when x = 2]
= f'(– 1) x f'(2)
= (– 3)(– 5)
= 15

(D) `"d"/"dx"[g(g(x))]`

= `g'(g(x))."d"/"dx"[g(x)]`
= g'(g(x)) x g'(x)

∴ `"d"/"dx"[g(g(x))]`at x = 2
= g'(g(2)) x g'(2)
= g'(– 1) c g'(2)    ...[∵ g(x) = – 1at x = 2]
= 4(– 4)
= – 16
Hence, (A) →3, (B) → 5, (C) → 4, (D) → 1.