We just found out that the derivative of \( x^2 + 1 \) was \( 2 x \) by doing a lot of algebra.

The good news is that we can look for patterns in how that algebra works, and find shortcuts to make finding derivatives much easier.

We'll start with the simplest shortcut:

If f is a constant function, the derivative is zero.

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ f( x + d ) - f( x ) } \over d } $$

No matter what x is, the value of f(x) is just a constant:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ c - c } \over d } $$

Subtract that constant from itself and we get zero in the numerator, and thus zero.

$$ f'(x) = {\lim_{ d \rightarrow 0 }} { 0 } $$

Zero is a constant. The limit of a constant is zero:

$$ f'(x) =  0 $$

The next shortcut is also about constants:

A constant factor can be moved through the derivative.

$$ f'(cf(x)) = c f'(f(x)) $$

The derivative of a sum is the sum of the derivatives.

$$ f'(f(x) + g(x)) = f'(f(x)) + f'(g(x)) $$

This also works for subtraction.

The next shortcut we demonstrated on the previous page. It is called the Power Rule.

$$ f'(x^n) = nx^{n-1} $$

To see how these simple rules help us, let's put them to use to find the derivative of \( x^2 + 1 \). All that algebra we did on the previous page just goes away:

The derivative of a sum is the sum of the derivatives:

$$ f'(x^2 + 1) = f'(x^2) + f'(1)$$

The derivative of a constant function is zero:

$$ f'(x^2 + 1)  =  f'(x^2) + 0$$

The Power Rule:

$$ f'(x^2 + 1)  =  f'(x^2)  =  2 x^{0} $$

Simplify:

$$ f'(x^2 + 1)  =  2 x^{0}  =  2 x$$

And we have our result:

$$ f'(x^2 + 1)  =  2 x$$