# Verify if f'(x)=g(x) . f(x)=x^3+ln(x), g(x)=1/x+3x^2.

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### 4 Answers

Given the functions:

f(x) = x^3 + lnx and g(x) = 1/x + 3x^2

We need to verify if f'(x) = g(x).

First , we need to differentiate f(x).

We will use the differentiation identities to find the derivative

We know that:

f'(x) = (x^3 + lnx)'

==> f'(x) = (x^3)' + (lnx)'

= 3x^2 + 1/x

We notice that:

f'(x) = 3x^2 + 1/x = g(x).

**Then, f'(x) = g(x).**

The question gives f(x)=x^3+ln(x) and g(x)=1/x+3x^2.

Differentiating f(x), we use the relations that the derivative of x^n is n*x^(n-1) and the derivative of ln x is 1/x.

These five f'(x) = 3*x^2 + 1/x

This is the same as g(x).

**Therefore f'(x) = g(x).**

To verify if f'(x)=g(x) . f(x)=x^3+ln(x), g(x)=1/x+3x^2.

f(x) = x^3 +ln(x).

We differentiate f(x) = x^3+lnx with respect to x.

f'(x) = (x^3+lnx)'

f'(x) = (x^3)' +(lnx)'

f'(x) = 3x^2+1/x.

Given g(x) = 1/x+3x^2 = 3x^2+1/x .

We now notice that f'(x) =g(x) = 3x^2+1/x.

We'll calculate the first derivative of f(x).

f'(x) = (x^3+lnx)'

f'(x) = 3x^2 + 1/x

The result of differentiating the function f(x) represents the expression of the function g(x) = 3x^2 + 1/x.

We could also prove that f(x) is the antiderivative of g(x) is calculating the indefinite integral of g(x), we'll get the expression of f(x).

Int g(x)dx = f(x) + C

Int (3x^2 + 1/x)dx = f(x) + C

We'll apply the property of the integral to be additive:

Int (3x^2 + 1/x)dx = Int (3x^2)dx + Int dx/x

Int (3x^2)dx = 3Int x^2dx

3Int x^2dx = 3*x^3/3 + C

We'll simplify and we'll get:

3Int x^2dx = x^3 + C (1)

Int dx/x = ln x + C (2)

Int g(x)dx = (1) + (2)

**Int g(x)dx = x^3 + ln x + C = f(x) + 0 = f(x)**

We notice that the result of the indefinite integral of g(x) is the function f(x) = x^3 + ln x (where C = 0).