Question

For the function Prove that

Answer 1

Let the given function be:

f( x )= x 100 100 + x 99 99 +…+ x 2 2 +x+1

We have to prove that

f ' ( 1 )=100 f ' ( 0 )

For a polynomial function

f( x )= a n x n + a n−1 x n−1 +… a 1 x+ a 0

Then,

d( f ) dx =n a n x n−1 +( n−1 ) a n−1 x n−2 +…+2 a 2 x+ a 1 (1)

Using the above formula for derivative we can now simplify the function:

df( x ) dx = d dx ( x 100 100 + x 99 99 +…+ x 2 2 +x+1 ) = d dx ( x 100 100 )+ d dx ( x 99 99 )+…+ d dx ( x 2 2 )+ d dx ( x )+ d dx ( 1 )

Using the theorem from equation 1:

d dx f( x )= 100 x 99 100 + 99 x 98 99 +…+ 2x 2 +1+0 = x 99 + x 98 + x 97 +…+x+1 (2)

We know that f ' ( x )= d dx f( x )

So, from equation (2)

f ' ( 0 )=0+0+⋯+0+1 =1

And f ' ( 1 )= 1 99+ 1 98 + 1 98 +…+1+1 =1⋅( 100 ) =100 =100⋅ f ' ( 0 ) (Count up to 100)

Or we can say that f ' ( 1 )=100 f ' ( 0 )

Hence, the expression has been proved.