f(x) = 2x^5 - 3x^3 + 5x

First we will determine first derivative f'(x)

f'(x) = 2*5x^4 - 3*3x^2 + 5*1

= 10x^4 - 9x^2 + 5

Now we will determine f''(x) by differentiate f'(x).

f''(x) = 10^4x^3 - 9*2x

= 40x^3 - 18x

Now we will substiute with x=1

==> f''(1) = 40*1 - 18*1 = 22

**==> f''(1) = 22**

If f(x) = 2x^5 - 3x^3 + 5x then find f''(1)

f'(x) = (2x^5-3x^3+5x)'

f'(x) = (2x^5)'-(3x^3)+(5x)'

f'(x) = 2*5x^(5-1) - 3*3*x^(3-1)+5x^(1-1), as (kx^5)' = k n x^(n-1)

f'(x) = 10x^4-9x^2 +5

f"(x) = (10x^4-9x^2+5)'

f"(x) = (10x^4)' - (9x^2) +(5)'

f"(x) = 10*4x^(4-1) -9*2x^(2-1)+0.

f"(x) = 40x^3-18x

f"(1) = 40(1)^3 -18*1

f"(x) = 40x^3-18x

f"(1) = 40-18 =22

We are given f(x) = 2x^5 - 3x^3 + 5x.

Now we use the relation that the derivative of x^n is nx^(n-1)

f(x) = 2x^5 - 3x^3 + 5x

=> f'(x) = 5 * 2* x^4 - 3*3* x^ 2 + 5

=> 10x^4 - 9 x^2 +5

f''(x) = 40 x^ 3 - 18 x

f''(1) = 40 * 1^3 - 18 * 1

= 40 - 18

= 22

**f''(1) is equal to 22**