Calculus

Differentiation of Functions

Higher-Order Derivatives

Page 1
Problems 1-3
Page 2
Problems 4-17

Higher-Order Derivatives of an Explicit Function

Let the function \(y = f\left( x \right)\) have a finite derivative \(f’\left( x \right)\) in a certain interval \(\left( {a,b} \right),\) i.e. the derivative \(f’\left( x \right)\) is also a function in this interval. If this function is differentiable, we can find the second derivative of the original function \(f\left( x \right)\), which is denoted as
\[{f^{\prime\prime} = f’ = {\left( {\frac{{dy}}{{dx}}} \right)^\prime } }={ \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) }={ \frac{{{d^2}y}}{{d{x^2}}}.}\] Similarly, if \(f”\) exists and is differentiable, we can calculate the third derivative of the function \(f\left( x \right)\):
\[{f^{\prime\prime\prime} = \frac{{{d^3}y}}{{d{x^3}}} }={ y^{\prime\prime\prime}.}\] The derivatives of higher order (if they exist) are defined as
\[
{{{f^{\left( 4 \right)}} = \frac{{{d^4}y}}{{d{x^4}}} }={ {y^{\left( 4 \right)}} = {\left( {{f^{\left( 3 \right)}}} \right)^\prime },} \ldots ,}\kern-0.3pt
{{f^{\left( n \right)}} = \frac{{{d^n}y}}{{d{x^n}}} = {y^{\left( n \right)}} }={ {\left( {{f^{\left( {n – 1} \right)}}} \right)^\prime }.}
\] Thus, the notion of the \(n\)th order derivative is introduced inductively by sequential calculation of \(n\) derivatives starting from the first order derivative. Transition to the next higher-order derivative is performed using the recurrence formula
\[{y^{\left( n \right)}} = {\left( {{y^{\left( {n – 1} \right)}}} \right)^\prime }.\] In some cases, we can derive a general formula for the derivative of an arbitrary \(n\)th order without computing intermediate derivatives. Some examples are considered below.

Note that the following linear relationships can be used for finding higher-order derivatives:
\[
{{\left( {u + v} \right)^{\left( n \right)}} = {u^{\left( n \right)}} + {v^{\left( n \right)}},\;\;\;}\kern-0.3pt
{{\left( {Cu} \right)^{\left( n \right)}} = C{u^{\left( n \right)}},\;\;}\kern-0.3pt{C = \text{const}.}
\]

Higher-Order Derivatives of an Implicit Function

The \(n\)th order derivative of an implicit function can be found by sequential (\(n\) times) differentiation of the equation \(F\left( {x,y} \right) = 0.\) At each step, after appropriate substitutions and transformations, we can obtain an explicit expression for the derivative, which depends only on the variables \(x\) and \(y\), i.e. the derivatives have the form
\[
{y’ = {f_1}\left( {x,y} \right),\;\;\;}\kern-0.3pt
{y^{\prime\prime} = {f_2}\left( {x,y} \right), \ldots,\;\;\;}\kern-0.3pt
{{y^{\left( n \right)}} = {f_n}\left( {x,y} \right).}
\]

Higher-Order Derivatives of a Parametric Function

Consider a function \(y = f\left( x \right)\) given parametrically by the equations
\[
\left\{
\begin{aligned}
x &= x\left( t \right) \\
y &= y\left( t \right)
\end{aligned}
\right..
\] The first derivative of this function is given by
\[y’ = {y’_x} = \frac{{{y’_t}}}{{{x’_t}}}.\] Differentiating once more with respect to \(x\), we find the second derivative:
\[y^{\prime\prime} = {y^{\prime\prime}_{xx}} = \frac{{{\left( {{y’_x}} \right)}’_t}}{{{x’_t}}}.\] Similarly, we define the derivatives of the third and higher order:
\[
{{y^{\prime\prime\prime} = {y^{\prime\prime\prime}_{xxx}} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}’_t}}}{{{x’_t}}}, \ldots,}\;\;}\kern0pt
{{{y^{\left( n \right)}} = y_{\underbrace {xx \ldots x}_n}^{\left( n \right)} }={ \frac{{{{\left( {y_{\underbrace {xx \ldots x}_{n – 1}}^{\left( {n – 1} \right)}} \right)}’_t}}}{{{x’_t}}}.}}
\]

Solved Problems

Click on problem description to see solution.

 Example 1

Find \(y^{\prime\prime}\), if \(y = x\ln x.\)

 Example 2

Find the second derivative of the function \(y = \sqrt[\large 4\normalsize]{{x + 1}}.\)

 Example 3

Calculate \(y^{\prime\prime}\) for the parabola equation \({y^2} = 4x.\)

 Example 4

Given the function \(y = {\left( {2x + 1} \right)^3}\left( {x – 1} \right).\) Find all derivatives of the \(n\)th order from \(n = 1\) to \(n = 5.\)

 Example 5

Find the \(n\)th order derivative of the natural logarithm function \(y = \ln x.\)

 Example 6

Find all derivatives of the sine function.

 Example 7

Find all derivatives of the cosine function.

 Example 8

Find all derivatives of the function \(y = {\large\frac{1}{x}\normalsize}.\)

 Example 9

Determine the second derivative of the function \(y = \arcsin {\large\frac{{{x^2} – 1}}{{{x^2} + 1}}\normalsize}.\)

 Example 10

Find the second derivative of the implicitly defined function \({x^2} + {y^2} = {R^2}\) (canonical equation of a circle).

 Example 11

Find the \(n\)th order derivative of the function \(y = {3^{2x + 1}}.\)

 Example 12

Find the \(n\)th order derivative of the power function \(y = {x^m}\) where \(m\) is a real number.

 Example 13

Find the \(n\)th order derivative of the square root \(y = \sqrt x .\)

 Example 14

Find the \(n\)th order derivative of the cube root \(y = \sqrt[\large 3\normalsize]{x}.\)

 Example 15

Given the equation of an ellipse in parametric form:
\[{x = a\cos t,\;\;\;}\kern-0.3pt{y = b\sin t,}\] where \(a\), \(b\) are semi-axes of the ellipse, \(t\) is a parameter. Find the first, second and third derivatives of the function \(y\) with respect to \(x.\)

 Example 16

Find the third derivative of the function given by the equation \({x^2} + 3xy + {y^2} = 1.\)

 Example 17

Find the second derivative of the function given by the equation \(x + y = {e^{x – y}}.\)

Example 1.

Find \(y^{\prime\prime}\), if \(y = x\ln x.\)

Solution.

Calculate the first derivative using the product rule:
\[
{y’ = \left( {x\ln x} \right)’ }
={ x’ \cdot \ln x + x \cdot {\left( {\ln x} \right)^\prime } }
={ 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1.}
\] Now we can find the second derivative:
\[
{y^{\prime\prime} = {\left( {\ln x + 1} \right)^\prime } }
= {\frac{1}{x} + 0 = \frac{1}{x}.}
\]

Example 2.

Find the second derivative of the function \(y = \sqrt[\large 4\normalsize]{{x + 1}}.\)

Solution.

We start with first derivative:
\[
y’ = {\left( {\sqrt[\large 4\normalsize]{{x + 1}}} \right)^\prime }
= {\left[ {{{\left( {x + 1} \right)}^{\large\frac{1}{4}\normalsize}}} \right]^\prime }
= {\frac{1}{4}{\left( {x + 1} \right)^{ – \large\frac{3}{4}\normalsize}} }={ \frac{1}{{4\sqrt[\large 4\normalsize]{{{{\left( {x + 1} \right)}^3}}}}}.}
\] Differentiate again to find the second derivative:
\[
{y^{\prime\prime} = {\left( {y’} \right)^\prime } = {\left[ {\frac{1}{4}{{\left( {x + 1} \right)}^{ – \large\frac{3}{4}\normalsize}}} \right]^\prime } }
= {\frac{1}{4} \cdot \left( { – \frac{3}{4}} \right){\left( {x + 1} \right)^{ – \large\frac{7}{4}\normalsize}} }
= { – \frac{3}{{16}} \cdot \frac{1}{{{{\left( {x + 1} \right)}^{\large\frac{7}{4}\normalsize}}}} }
= { – \frac{3}{{16\sqrt[\large 4\normalsize]{{{{\left( {x + 1} \right)}^7}}}}}.}
\]

Example 3.

Calculate \(y^{\prime\prime}\) for the parabola equation \({y^2} = 4x.\)

Solution.

By implicit differentiation,
\[{2yy’ = 4,\;\; }\Rightarrow {yy’ = 2.}\] Differentiating again and using the product rule, we obtain
\[y’y’ + yy^{\prime\prime} = 0.\] Multiply both sides by \({y^2}\):
\[{y^2}{\left( {y’} \right)^2} + {y^3}y^{\prime\prime} = 0.\] As \(yy’ = 2\) and, hence, \({\left( {yy’} \right)^2} = 4,\) we can write the last equation as
\[4 + {y^3}y^{\prime\prime} = 0.\] Then
\[y^{\prime\prime} = – \frac{4}{{{y^3}}}.\]

Page 1
Problems 1-3
Page 2
Problems 4-17