Question

f(x) =x¹⁰⁰⇒f'(1) is equal to

Answer 1

Answer:

• f(x)=x100 + x99 + ………….x + 1

• f'(x) = 100x99 + 99x98 +…… + 1

• f'(1) = 100(1)99 + 99(1)98 + …… + 1

[Using Arithmetic Progression, where d = -1, a = 100 & n = 100] = 50 (200 – 99) = 50 (101) = 5050

Answer 2

Step-by-step explanation:

We need to find the derivative of f(x) at x=1

i.e. f

′

(1)

f(x)=1+x+x

2

+......+x

100

⇒f

′

(x)=(1+x+x

2

+.....+x

100

)

′

⇒f

′

(x)=0+1+2x+.....+100x

99

∴f

′

(x)=1+2x+.....+100x

99

Putting x=1 in f

′

(x)

⇒f

′

(1)=1+2×1+.....+100×1

99

⇒f

′

(1)=1+2+....+100

⇒f

′

(1)=

2

100(100+1)

= 5050

⇒f

′

(1) = 5050

please follow me