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# 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.

$y’ = {\left( {\sqrt[\large 4\normalsize]{{x + 1}}} \right)^\prime } = {\left[ {{{\left( {x + 1} \right)}^{\large\frac{1}{4}\normalsize}}} \right]^\prime } = {\frac{1}{4}{\left( {x + 1} \right)^{ – \large\frac{3}{4}\normalsize}} }={ \frac{1}{{4\sqrt[\large 4\normalsize]{{{{\left( {x + 1} \right)}^3}}}}}.}$ 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