# Calculus

## Differentiation of Functions # Second Derivatives

### Second Derivative 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 $$f\left( x \right).$$

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

${\frac{{{d^2}f}}{{d{x^2}}}\;\text{ or }\;\frac{{{d^2}y}}{{d{x^2}}}\;}\kern0pt{\left( \text{Leibniz’s notation} \right)}$

${f^{\prime\prime}\left( x \right)\;\text{ or }\;y^{\prime\prime}\left( x \right)\;}\kern0pt{\left( \text{Lagrange’s notation} \right)}$

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:

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

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

The second derivatives satisfy the following linear relationships:

${{\left( {u + v} \right)^{\prime\prime}} = {u^{\prime\prime}} + {v^{\prime\prime}},\;\;\;}\kern-0.3pt {{\left( {Cu} \right)^{\prime\prime}} = C{u^{\prime\prime}},\;\;}\kern-0.3pt{C = \text{const}.}$

### 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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

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

### Example 6

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

### Example 7

Find $$y^{\prime\prime},$$ if $$y = x\sqrt {1 + {x^2}}.$$

### Example 8

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

### Example 9

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

### Example 10

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

### Example 11

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

### Example 12

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

### Example 13

The function $$y = f\left( x \right)$$ is given in parametric form by the equations $x = {t^3},\;\;y = {t^2} + 1,$ where $$t \gt 0.$$ Find $$y_{xx}^{\prime\prime}.$$

### Example 14

Find the second derivative of the function given by the equation $${x^3} + {y^3} = 1.$$

### Example 15

The function $$y = f\left( x \right)$$ is given in parametric form by the equations${x = t + \cos t,\;\;}\kern0pt{y = 1 + \sin t,}$where $$t \in \left( {0,2\pi } \right).$$ Find $$y_{xx}^{\prime\prime}.$$

### Example 16

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

### Example 1.

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

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:

${y^{\prime\prime} = {\left( {\ln x + 1} \right)^\prime } } = {\frac{1}{x} + 0 = \frac{1}{x}.}$

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Problems 1-5
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Problems 6-16