# The curve with equation y= x^5 + 20x^2 - 8 passes through the point P, where x = -2. Find the value of (d^2)y / dx^2 at the point P.

### 2 Answers | Add Yours

You need to diferentiate the function with respect to x such that:

`f'(x) = (dy)/(dx) = 5x^4 + 40x`

You need to differentiate the function `(dy)/(dx) = 5x^4 + 40x` with respect to x such that:

`f"(x) = (d^2y)/(dx^2) = 20x^3 + 40`

You need to substitute -2 for x in equation `(d^2y)/(dx^2) = 20x^3 + 40` such that:

`f"(-2) = 20*(-2)^3 + 40`

`f"(-2) = - 160 + 40 =gt f"(-2) = -120`

**Hence, evaluating `(d^2y)/(dx^2)` at `x = -` 2 yields `f"(-2) = -120.` **

dy/dx = 5x^4 + 40x

(d^2)y / dx^2 = 20x^3 +40

Sub x= -2,

(d^2)y / dx^2

= 20(-2)^3 +40

= -120 :)