### Differentiation: Applications of derivatives

### Higher order derivatives

The calculation of derivatives is not limited to the first and the second derivative.

We write #f^{(n)}#, in which #n# is an integer, for the #n#-th derivative of #f#.

Therefore,

\[\begin{array}{rcl}f^{(0)}&=&f\\f^{(1)}&=&f'\\f^{(2)}&=&f''\\\cdots\\f^{(n)}&=&f''''^{...}\end{array}\]

Here, the last expression consists of #n# times an apostrophe.

We determine #f^{(n)}# by differentiating the function #f# exactly #n# times.

**Example**

\[\begin{array}{rcl} f(x)&=&x^7 \\ \\ f^{(1)}(x)&=&7x^6 \\ \\f^{(2)}(x)&=&42x^5 \\ \\f^{(3)} (x)&=&210x^4\\ \\f^{(4)}(x)&=&840x^3 \\ &\vdots& \\f^{(8)}(x)&=&0 \end{array}\]

Calculate #f'(x)#, #f''(x)#, and #f'''(x)# for \[f(x)=5\cdot x^6-4\cdot x^2-5\cdot x-4\tiny.\]

#f'(x)=# #30\cdot x^5-8\cdot x-5#

#f''(x)=# #150\cdot x^4-8#

#f'''(x)=# #600\cdot x^3#

First we determine #f'(x)# using the power rule. This gives:

\[f'(x)=30\cdot x^5-8\cdot x-5\]

Next, we determine # f''(x)# by differentiating #f'(x)# in the same way. This gives:

\[f''(x)=150\cdot x^4-8\]

Finally, we determine #f'''(x)# by differentiating #f''(x)# in the same way. This gives:

\[f'''(x)=600\cdot x^3\]

#f''(x)=# #150\cdot x^4-8#

#f'''(x)=# #600\cdot x^3#

First we determine #f'(x)# using the power rule. This gives:

\[f'(x)=30\cdot x^5-8\cdot x-5\]

Next, we determine # f''(x)# by differentiating #f'(x)# in the same way. This gives:

\[f''(x)=150\cdot x^4-8\]

Finally, we determine #f'''(x)# by differentiating #f''(x)# in the same way. This gives:

\[f'''(x)=600\cdot x^3\]

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