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.
