1. function of a derivative

find a function whose derivative is 20x^3 plus 6x^2 plus 5

2. Originally Posted by eduk8tedidiot
find a function whose derivative is 20x^3 plus 6x^2 plus 5

What you want to find here is the 'anti-derivative' of the function. The integral of a derivative gives you the antiderivative.

Hence, if we are seeking function $\displaystyle f(x)$, then

$\displaystyle f(x) = \int 20x^3 + 6x^2 + 5 \, dx$

3. Originally Posted by eduk8tedidiot
find a function whose derivative is 20x^3 plus 6x^2 plus 5

Have you heard of an integral? Well anyway, we're looking for the anti-derivative of $\displaystyle 20x^3+6x+62+5$

In order to get an $\displaystyle x^3$ the power rule tells us we need an $\displaystyle x^4$ but the derivative of $\displaystyle x^4=4x^3$ so we'd have a 4 out in front when we want a 20, so let's put a 5 out in front of our original guess

Now the derivative of $\displaystyle 5x^4=20x^3$ so the first term is done....

Doing similar steps will lead you to an answer of $\displaystyle 20x^4+2x^3+5x+c$

Note that when we take the derivative of that, the +c has a derivative of zero, since the question is to find an antiderivative, you can choose any value for c as the answer

4. $\displaystyle f'(x) = 20x^3 + 6x^2 + 5$

$\displaystyle f(x) = 5x^4 + 2x^3 + 5x + C$

$\displaystyle C = konst$

$\displaystyle f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$

$\displaystyle f'(x) = a_1 + 2a_2x + 3a_3x^2 + ...+ na_nx^{n-1}$

$\displaystyle f(x) = C$ C = const. $\displaystyle \rightarrow f'(x) = 0$

$\displaystyle f(x) = kx^m \rightarrow f'(x) = m*k*x^{m-1}$

5. Originally Posted by josipive
$\displaystyle f'(x) = 20x^3 + 6x^2 + 5$

$\displaystyle f(x) = 5x^4 + 2x^3 + 5x + C$

$\displaystyle C = konst$

$\displaystyle f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$

$\displaystyle f'(x) = a_1 + 2a_2x + 3a_3x^2 + ...+ na_nx^{n-1}$

$\displaystyle f(x) = C$ C = const. $\displaystyle \rightarrow f'(x) = 0$

$\displaystyle f(x) = kx^m \rightarrow f'(x) = m*k*x^{m-1}$