# Second Derivatives

## Second Derivative of an Explicit Function

Let the function f (x) have a finite derivative f '(x) in a certain interval (a, b), i.e. the derivative f '(x) is also a function in this interval. If this function is differentiable, we can find the second derivative of the original function f (x).

The second derivative (or the second order derivative) of the function f (x) may be denoted as

$\frac{{{d^2}f}}{{d{x^2}}}\;\;\text{or}\;\;\frac{{{d^2}y}}{{d{x^2}}}\;\left( \text{Leibniz's notation} \right)$
$f^{\prime\prime}\left( x \right)\;\;\text{or}\;\;y^{\prime\prime}\left( x \right)\;\left( \text{Lagrange's notation} \right)$

The second derivative has many applications. In particular, it can be used to determine the concavity and inflection points of a function as well as minimum and maximum points.

In physics, when we have a position function $$\mathbf{r}\left( t \right)$$, the first derivative is the velocity $$\mathbf{v}\left( t \right)$$ and the second derivative is the acceleration $$\mathbf{a}\left( t \right)$$ of the object:

$\mathbf{a}\left( t \right) = \frac{{d\mathbf{v}}}{{dt}} = \mathbf{v}^\prime\left( t \right) = \frac{{{d^2}\mathbf{r}}}{{d{t^2}}} = \mathbf{r}^{\prime\prime}\left( t \right).$

In physics, when we have a position function $$\mathbf{r}\left( t \right)$$, the first derivative is the velocity $$\mathbf{v}\left( t \right)$$ and the second derivative is the acceleration $$\mathbf{a}\left( t \right)$$ of the object:

Other applications of the second derivative are considered in chapter Applications of the Derivative.

The second derivatives satisfy the following linear relationships:

${\left( {u + v} \right)^{\prime\prime}} = {u^{\prime\prime}} + {v^{\prime\prime}},\;\;\;{\left( {Cu} \right)^{\prime\prime}} = C{u^{\prime\prime}},\;\;C = \text{const}.$

## Second Derivative of an Implicit Function

The second derivative of an implicit function can be found using sequential differentiation of the initial equation $$F\left( {x,y} \right) = 0.$$ At the first step, we get the first derivative in the form $$y^\prime = {f_1}\left( {x,y} \right).$$ On the next step, we find the second derivative, which can be expressed in terms of the variables $$x$$ and $$y$$ as $$y^{\prime\prime} = {f_2}\left( {x,y} \right).$$

## Second Derivative of a Parametric Function

Consider a parametric function $$y = f\left( x \right)$$ given 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

$y' = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}}.$

Differentiating once more with respect to $$x,$$ we find the second derivative:

$y^{\prime\prime} = {y^{\prime\prime}_{xx}} = \frac{{{\left( {{y'_x}} \right)}'_t}}{{{x'_t}}}.$

## Solved Problems

### Example 1.

Find the second derivative of the function $y = \frac{x}{{\sqrt {1 - {x^2}} }}.$

Solution.

By the quotient and chain rules, we get

$y^\prime = \left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)^\prime = \frac{{x^\prime\sqrt {1 - {x^2}} - x\left( {\sqrt {1 - {x^2}} } \right)^\prime}}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}} = \frac{{1 \cdot \sqrt {1 - {x^2}} - x \cdot \frac{{\left( { - 2x} \right)}}{{2\sqrt {1 - {x^2}} }}}}{{1 - {x^2}}} = \frac{{\sqrt {1 - {x^2}} + \frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}}}{{1 - {x^2}}} = \frac{{\frac{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2} + {x^2}}}{{\sqrt {1 - {x^2}} }}}}{{1 - {x^2}}} = \frac{{1 - {x^2} + {x^2}}}{{\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }} = \frac{1}{{\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }}.$

Differentiate again using the power and chain rules:

$y^{\prime\prime} = \left( {\frac{1}{{\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }}} \right)^\prime = \left( {{{\left( {1 - {x^2}} \right)}^{ - \frac{3}{2}}}} \right)^\prime = - \frac{3}{2}{\left( {1 - {x^2}} \right)^{ - \frac{5}{2}}} \cdot \left( { - 2x} \right) = \frac{{3x}}{{{{\left( {1 - {x^2}} \right)}^{\frac{5}{2}}}}} = \frac{{3x}}{{\sqrt {{{\left( {1 - {x^2}} \right)}^5}} }}.$

### Example 2.

Find the second derivative of the polynomial function $y = 3{x^4} - 2{x^3} + 4{x^2} - 5x + 1.$

Solution.

Take the first derivative using the power rule and the basic differentiation rules:

$y^\prime = 12{x^3} - 6{x^2} + 8x - 5.$

Differentiate once more to find the second derivative:

$y^{\prime\prime} = 36{x^2} - 12x + 8.$

### Example 3.

Find the second derivative of the polynomial function $y = 2{x^5} + 3{x^4} - 4{x^3} + {x^2} - 6.$

Solution.

Take the first derivative using the power rule and the basic differentiation rules:

$y^\prime = 10{x^4} + 12{x^3} - 12{x^2} + 2x.$

The second derivative is expressed in the form

$y^{\prime\prime} = 40{x^3} + 36{x^2} - 24x + 2.$

### Example 4.

Find $$y^{\prime\prime},$$ if $y = \cot x.$

Solution.

The first derivative of the cotangent function is given by

$y^\prime = \left( {\cot x} \right)^\prime = - \frac{1}{{{{\sin }^2}x}}.$

Differentiate it again using the power and chain rules:

$y^{\prime\prime} = \left( { - \frac{1}{{{{\sin }^2}x}}} \right)^\prime = - \left( {{{\left( {\sin x} \right)}^{ - 2}}} \right)^\prime = \left( { - 1} \right) \cdot \left( { - 2} \right) \cdot {\left( {\sin x} \right)^{ - 3}} \cdot \left( {\sin x} \right)^\prime = \frac{2}{{{{\sin }^3}x}} \cdot \cos x = \frac{{2\cos x}}{{{{\sin }^3}x}}.$

### Example 5.

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 6.

Find $$y^{\prime\prime},$$ if $y = {e^{ - {x^2}}}.$

Solution.

Take the first derivative:

$y^\prime = \left( {{e^{ - {x^2}}}} \right)^\prime = {e^{ - {x^2}}} \cdot \left( { - {x^2}} \right)^\prime = - 2x{e^{ - {x^2}}}.$

Then the second derivative is given by

$y^{\prime\prime} = \left( { - 2x{e^{ - {x^2}}}} \right)^\prime = \left( { - 2x} \right)^\prime{e^{ - {x^2}}} - 2x\left( {{e^{ - {x^2}}}} \right)^\prime = - 2{e^{ - {x^2}}} - 2x \cdot \left( { - 2x{e^{ - {x^2}}}} \right) = - 2{e^{ - {x^2}}} + 4{x^2}{e^{ - {x^2}}} = {e^{ - {x^2}}}\left( {4{x^2} - 2} \right).$

See more problems on Page 2.