# `f(x) = e^x - x^3` Find the first and second derivatives of the function. Check to see that your answers aer reasonable by comparing the graphs of f, f', and f''

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### 4 Answers

Given function is:

`f(x)=e^x-x^3`

Now we will find the first derivative:

`f'(x)=d/dx(e^x-x^3)`

`=d/dx(e^x)-d/dx(x^3)`

`=e^x-3x^2`

The second derivative is obtained by taking the derivative of the first derivative.

`f''(x)=d/dx(f'(x))=d/dx(e^x-3x^2)`

`=d/dx(e^x)-d/dx(3x^2)`

`=e^x-6x`

Comparison of graphs of f,f' and f'' is shown below:

`f(x)=e^x-x^3`

`f'(x)=d/dx(e^x-x^3)`

`f'(x)=d/dx(e^x)-d/dx(x^3)`

`f'(x)=e^x-3x^2`

now we will find 2nd derivative

`f''(x)=d/dx(e^x)-d/dx(3x^2)`

`f''(X)=e^x-3*2(x)`

`f''(x)=e^x-6x`

**Note:- **

**1) If f(x) = e^(nx) ; where n = a real number;**

** then f'(x) = n*e^(nx)**

**2) If f(x) = x^n ; where n = a real number**

**then f'(x) = n*x^(n-1)**

Now, given f(x) = (e^x) - (x^3)

Thus, f'(x) = (e^x) - 3*(x^2)

f"(x) = (e^x) - 6*(x^1)

or, f"(x) = (e^x) - 6x

F(x)=e^x-x^3

F'(x)=e^x-3x^2

F''(x)=e^x-3.2x^1