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.

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:

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

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:

\[y^{\prime\prime} = 36{x^2} – 12x + 8.\]

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

\[y^{\prime\prime} = 40{x^3} + 36{x^2} – 24x + 2.\]

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:

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

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:

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