# Basic differentiation review

Review the basic differentiation rules and use them to solve problems.

## What are the basic differentiation rules?

Sum rulestart fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, plus, g, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis, plus, start fraction, d, divided by, d, x, end fraction, g, left parenthesis, x, right parenthesis
Difference rulestart fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, minus, g, left parenthesis, x, right parenthesis, close bracket, equals, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis, minus, start fraction, d, divided by, d, x, end fraction, g, left parenthesis, x, right parenthesis
Constant multiple rulestart fraction, d, divided by, d, x, end fraction, open bracket, k, dot, f, left parenthesis, x, right parenthesis, close bracket, equals, k, dot, start fraction, d, divided by, d, x, end fraction, f, left parenthesis, x, right parenthesis
Constant rulestart fraction, d, divided by, d, x, end fraction, k, equals, 0
The Sum rule says the derivative of a sum of functions is the sum of their derivatives.
The Difference rule says the derivative of a difference of functions is the difference of their derivatives.
The Constant multiple rule says the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
The Constant rule says the derivative of any constant function is always 0.

## What problems can I solve with basic differentiation rules?

You can find the derivatives of functions that are combinations of other, simpler, functions. For example, H, left parenthesis, x, right parenthesis is defined as 2, f, left parenthesis, x, right parenthesis, minus, 3, g, left parenthesis, x, right parenthesis, plus, 5. We can find H, prime, left parenthesis, x, right parenthesis as follows;
\begin{aligned} &\phantom{=}H'(x) \\\\ &=\dfrac{d}{dx}H(x)&&\gray{\text{Equivalent notation}} \\\\ &=\dfrac{d}{dx}[2f(x)-3g(x)+5]&&\gray{\text{Substitute the expression for }H(x)} \\\\ &=\dfrac{d}{dx}[2f(x)]-\dfrac{d}{dx}[3g(x)]+\dfrac{d}{dx}(5)&&\gray{\text{Sum and difference rules}} \\\\ &=2f'(x)-3g'(x)+0&&\gray{\text{Constant and constant multiple rules}} \end{aligned}
We used the basic differentiation rules to find that H, prime, left parenthesis, x, right parenthesis, equals, 2, f, prime, left parenthesis, x, right parenthesis, minus, 3, g, prime, left parenthesis, x, right parenthesis.
Now suppose we are also given that start color blueD, f, prime, left parenthesis, 3, right parenthesis, equals, 1, end color blueD and start color goldD, g, prime, left parenthesis, 3, right parenthesis, equals, 5, end color goldD. We can find H, prime, left parenthesis, 3, right parenthesis as follows:
\begin{aligned} H'(3)&=2\blueD{f'(3)}-3\goldD{g'(3)} \\\\ &=2(\blueD1)-3(\goldD5) \\\\ &=-13 \end{aligned}