The operation of differentiation or finding the derivative of a function has the fundamental property of linearity. This property makes taking the derivative easier for functions constructed from the basic elementary functions using the operations of addition and multiplication by a constant number. The basic differentiation rules allow us to compute the derivatives of such functions without using the formal definition of the derivative. Consider these rules in more detail.

Derivative of a Constant

If \(f\left( x \right) = C,\) then

\[f’\left( x \right) = C’ = 0.\]

The proof of this rule is considered on the Definition of the Derivative page.

Constant Multiple Rule

Let \(k\) be a constant. If \(f\left( x \right)\) is differentiable, then \(kf\left( x \right)\) is also differentiable and

\[{\left( {kf\left( x \right)} \right)^\prime } = kf’\left( x \right).\]

Sum Rule

Let \(f\left( x \right)\) and \(g\left( x \right)\) be differentiable functions. Then the sum of two functions is also differentiable and

\[{\left( {f\left( x \right) + g\left( x \right)} \right)^\prime } ={ f’\left( x \right) + g’\left( x \right).}\]

Let \(n\) functions \({f_1}\left( x \right)\), \({f_2}\left( x \right)\), \(\ldots\), \({f_n}\left( x \right)\) be differentiable. Then their sum is also differentiable and

\[
{{\left[ {{f_1}\left( x \right) + {f_2}\left( x \right) + \ldots + {f_n}\left( x \right)} \right]^\prime } }
= {{f_1}^\prime \left( x \right) + {f_2}^\prime \left( x \right) + \ldots + {f_n}^\prime \left( x \right).}
\]

Combining the both rules we see that the derivative of difference of two functions is equal to the difference of the derivatives of these functions assuming both of the functions are differentiable:

\[{\left( {f\left( x \right) – g\left( x \right)} \right)^\prime } = {f’\left( x \right) – g’\left( x \right).}\]

We can write the common rule:

Linear Combination Rule

Suppose \(f\left( x \right)\) and \(g\left( x \right)\) are differentiable functions and \(a,\) \(b\) are real numbers. Then the function \(h\left( x \right) = af\left( x \right) + bg\left( x \right)\) is also differentiable and

\[h’\left( x \right) = af’\left( x \right) + bg’\left( x \right).\]

We add to this list one more simple rule:

Derivative of the Function \(y = x\)

If \(f\left( x \right) = x,\) then

\[f’\left( x \right) = {\left( x \right)^\prime } = 1.\]

This formula is derived on the Definition of the derivative page.

Solved Problems

Click or tap a problem to see the solution.

Solution.

Using the linear differentiation rules, we have

\[
{y’\left( x \right) = {\left( {{x^2} – 5x} \right)^\prime } }
= {{\left( {{x^2}} \right)^\prime } – {\left( {5x} \right)^\prime } }
= {{\left( {{x^2}} \right)^\prime } – 5{\left( x \right)^\prime } }
= {2x – 5 \cdot 1 = 2x – 5.}
\]

Solution.

\[
{y’\left( x \right) = {\left( {\frac{{ax + b}}{{a + b}}} \right)^\prime } }
= {\frac{1}{{a + b}} \cdot {\left( {ax + b} \right)^\prime } = {\frac{a}{{a + b}}.}}
\]

Solution.

Using the basic differentiation rules, we obtain:

\[
{y’\left( x \right) = {\left( {2\sqrt x – 3\sin x} \right)^\prime } }
= {{\left( {2\sqrt x } \right)^\prime } – {\left( {3\sin x} \right)^\prime } }
= {2{\left( {\sqrt x } \right)^\prime } – 3{\left( {\sin x} \right)^\prime } }
= {2 \cdot \frac{1}{{2\sqrt x }} – 3\cos x }
= {\frac{1}{{\sqrt x }} – 3\cos x.}
\]

Solution.

This expression is a linear combination of two trigonometric functions. The derivative has the following form:

\[ {y’\left( x \right) = {\left( {3\sin x + 2\cos x} \right)^\prime } } = {{\left( {3\sin x} \right)^\prime } + {\left( {2\cos x} \right)^\prime } } = {3{\left( {\sin x} \right)^\prime } + 2{\left( {\cos x} \right)^\prime } } = {3 \cdot \cos x + 2 \cdot \left( { – \sin x} \right) } = {3\cos x – 2\sin x.} \]

Solution.

As \({3^2} – 8 = 1 \gt 0\), the function at the point \(x = 3\) is equivalent to

\[y\left( x \right) = x + {x^2} – 8.\]

So, by the sum rule and the power rule (see “Derivatives of Power Functions“),

\[ {y’\left( x \right) = {\left( {x + {x^2} – 8} \right)^\prime } } = {x’ + {\left( {{x^2}} \right)^\prime } – 8′ } = {1 + 2x + 0 }={ 2x + 1.} \]

At the point \(x = 3\) the value of the derivative is

\[y’\left( 3 \right) = 2 \cdot 3 + 1 = 7.\]