# If f(2)=10 and f'(x)=x^2f(x) for all x, find f”(2).

The aim of this question is to learn how to evaluate the values of a higher order derivative without explicitly declaring the function itself.

Derivative

To solve such problems, we may need to solve the basic rules of finding the derivatives. These include the power rule and product rule etc.

Power of derivative

According to the power rule of differentiation:

$\dfrac{ d }{ dx } \bigg ( x^{ n } \bigg ) \ = \ n \ x^{ n – 1 }$

Product of derivative

According to the product rule of differentiation:

$\dfrac{ d }{ dx } \bigg ( f ( x ) \ g ( x ) \bigg ) \ = \ f^{‘} (x) \ g ( x ) \ + \ f ( x ) \ g^{‘} ( x )$

Given:

$f^{‘} ( x ) \ = \ x^2 \ f ( x )$

Substitute $x \ = \ 2$ in the above equation:

$f^{‘} ( 2 ) \ = \ ( 2 )^{ 2 } f ( 2 )$

$f^{‘} ( 2 ) \ = \ 4 \ f ( 2 )$

Substitute $f(2) \ = \ 10$ in the above equation:

$f^{‘} ( 2 ) \ = \ 4 \ ( 10 )$

$f^{‘} ( 2 ) \ = \ 40$

Recall the given equation again:

$f^{‘} ( x ) \ = \ x^2 \ f ( x )$

Differentiating the above equation:

$\dfrac{ d }{ dx } \bigg ( f^{‘} ( x ) \bigg ) \ = \ \dfrac{ d }{ dx } \bigg ( x^{ 2 } f ( x ) \bigg )$

$f^{ ” } ( x ) \ = \ \dfrac{ d }{ dx } \bigg ( x^{ 2 } \bigg ) \ f ( x ) \ + \ x^{ 2 } \ \dfrac{ d }{ dx } \bigg ( f ( x ) \bigg )$

$f^{ ” } ( x ) \ = \ \bigg ( 2 x \bigg ) \ f(x) \ + \ x^{ 2 } \ \bigg ( f^{‘} ( x ) \bigg )$

$f^{ ” } ( x ) \ = \ 2 x \ f(x) \ + \ x^{ 2 } \ f^{‘} ( x )$

Substitute $x \ = \ 2$ in the above equation:

$f^{ ” } ( 2 ) \ = \ 2 (2) \ f(2) \ + \ ( 2 )^{ 2 } f^{‘} ( 2 )$

$f^{ ” } ( 2 ) \ = \ 4 f ( 2 ) \ + \ 4 f^{‘} ( 2 )$

Substitute $f ( 2 ) \ = \ 10$ and $f^{‘} ( 2 ) \ = \ 40$ in the above equation:

$f^{ ” } ( 2 ) \ = \ 4 (10) \ + \ 4 (40)$

$f^{ ” } ( 2 ) \ = \ 40 \ + \ 160$

$f^{ ” } ( 2 ) \ = \ 200$

## Numerical Result

$f^{ ” } ( 2 ) \ = \ 200$

## Example

Given that $f ( 10 ) \ = \ 1$ and $f^{‘} ( x ) \ = \ x f ( x )$, find the value of f^{ ” } ( 10 ) $. Given: $f^{‘} ( x ) \ = \ x \ f ( x )$ Substitute$ x \ = \ 10 $in the above equation: $f^{‘} ( 10 ) \ = \ ( 10 ) f ( 10 )$ Substitute$ f(10) \ = \ 1 $in the above equation: $f^{‘} ( 10 ) \ = \ 10 \ ( 1 )$ $f^{‘} ( 10 ) \ = \ 10$ Recall the given equation again: $f^{‘} ( x ) \ = \ x \ f ( x )$ Differentiating the above equation: $\dfrac{ d }{ dx } \bigg ( f^{‘} ( x ) \bigg ) \ = \ \dfrac{ d }{ dx } \bigg ( x f ( x ) \bigg )$ $f^{ ” } ( x ) \ = \ \dfrac{ d }{ dx } \bigg ( x \bigg ) \ f ( x ) \ + \ x \ \dfrac{ d }{ dx } \bigg ( f ( x ) \bigg )$ $f^{ ” } ( x ) \ = \ \bigg ( 1 \bigg ) \ f(x) \ + \ x \ \bigg ( f^{‘} ( x ) \bigg )$ $f^{ ” } ( x ) \ = \ f(x) \ + \ x \ f^{‘} ( x )$ Substitute$ x \ = \ 10 $in the above equation: $f^{ ” } ( 10 ) \ = \ f(10) \ + \ ( 10 ) f^{‘} ( 10 )$ Substitute$ f ( 10 ) \ = \ 1 $and$ f^{‘} ( 10 ) \ = \ 10 \$ in the above equation:

$f^{ ” } ( 10 ) \ = \ (1) \ + \ 10 (10)$

$f^{ ” } ( 10 ) \ = \ 1 \ + \ 100$

$f^{ ” } ( 10 ) \ = \ 101$