Derivative of Functions in Parametric Form: Definition, Second Order Derivative, Applications

Derivative of functions in parametric forms: We know that chain rule, product rule and quotient rule can be used to calculate the derivatives of standard functions. These rules can also be used to calculate the derivatives of complex functions. These functions have two variables that are related to each other in implicit and explicit functions. Sometimes, we come across a few functions in which the variables are not related to each other implicitly or explicitly, in fact, they are only related to each other over a third variable. The third variable is called a parameter. Such equations are known as parametric forms. In this article, let us discuss the derivatives of parametric equations.

First-Order Derivative

A parametric relation between \(x\) and \(y\) can be expressed as \(x = f\left( t \right)\) and \(y = g\left( t \right)\) is the parametric form representation with the parameter \(t\), such that \(y\) is defined as a function of \(x\).

The chain rule of derivatives is \(\frac{{dy}}{{dt}} = \frac{{dy}}{{dx}} \times \frac{{dx}}{{dt}}\). The chain rule can also be rewritten as,
\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\), where \(\frac{{dx}}{{dt}} \ne 0\)

As mentioned earlier, \({y^\prime } = {g^\prime }(t) = \frac{{dy}}{{dt}}\) and \({x^\prime } = {f^\prime }(t) = \frac{{dx}}{{dt}}\)

Thus, we can say \(\frac{{dy}}{{dx}} = \frac{{{g^\prime }(t)}}{{{f^\prime }(t)}}\), where \({f^\prime }(t) \ne 0\)

Now, let us see the rules/steps for solving problems on derivatives of functions expressed in the parametric form:
Step 1: Write the given functions \(x\) and \(y\) in terms of the other parameter \(t\).
Step 2: Find \(\frac{{dy}}{{dt}}\) and \(\frac{{dx}}{{dt}}\).
Step 3: Use the formula and solving functions on parametric form, i.e. \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)
Step 4: Substitute the values of \(\frac{{dy}}{{dt}}\) and \(\frac{{dx}}{{dt}}\) obtained from step \(3\).
Step 5: Simplify to get the final result.

Learn about Derivative of a Function Here

Applications of Parametric Functions

Parametric functions are mainly used in the integration of different types of functions if the given function is in complex form. So in such cases, parameter \(t\) is used for substitution for some part of the function.

There are two types of parametric equations that are typical in real-life situations. The first is circular motion. The second is projectile motion. The applications of parametric functions are listed below.

The parametric equations are generally used in Physics, to find the acceleration, i.e. \(a = \frac{{dv}}{{dt}}\)  or \(a = v\frac{{dv}}{{dx}}\)

Here, the functions \(v\) (velocity) and \(x\) (position) are usually expressed in terms of time \(t\) (parameter).

Let us see some examples for solving the functions of the parametric form.

Example 1: Find the value of \(\frac{{dy}}{{dx}}\), if \(x = a\,\cos \left( \theta  \right),\,y = a\,\sin \left( \theta  \right)\).
Solution: Given: \(x = a\,\cos \left( \theta  \right),\,y = a\,\sin \left( \theta  \right)\)

Now, differentiating \(x\) with respect to \(\theta \), we get,

\(\frac{{dx}}{{d\theta }} = – a\,\sin \left( \theta  \right)\)

Similarly, differentiating \(y\) with respect to \(\theta \), we get,

\(\frac{{dy}}{{d\theta }} = a\,\cos \left( \theta  \right)\)

Now, let’s find \(\frac{{dy}}{{dx}}\)

We know that,

\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{a\cos (\theta )}}{{ – a\sin (\theta )}}\)

\(\therefore \,\frac{{dy}}{{dx}} = – \cot (\theta )\)

Example 2: Find the value of \(\frac{{dy}}{{dx}}\), if \(x = a{t^2} + 2t,y = 2t\) at \(t = 0\).
Solution: Given: \(x = a{t^2} + 2t,y = 2t\)

Now, differentiating \(x\) with respect to \(t\), we get,

\(\frac{{dx}}{{dt}} = 2at+2\)

Similarly, differentiating \(y\) with respect to \(t\), we get

\(\frac{{dy}}{{dt}} = 2\)

Now, let’s find \(\frac{{dy}}{{dx}}\)

We know that,

\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{{2at + 2}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{at + 1}}\)

When \(t=0\), then \(\frac{{dy}}{{dx}} = 1\)

\(\therefore \,\frac{{dy}}{{dx}} = 1\)

Second-Order Derivative

In general, the second-order derivative is defined as the derivative of the first-order derivative of the given function. The first-order derivative at a given point is the same as the slope of the tangent at that point. Alternatively, it is also defined as the instantaneous rate of change of a function at that point. Similarly, the second-order derivative gives us the idea of the shape of the graph of a given function. The second -derivative of the function \(f(x)\) is generally denoted by \({f^{\prime \prime }}(x)\). It is also represented as \({D^2}y\) or \({y_2}\) or \({y^{\prime \prime }}\), where \(y = f(x)\).

The graphical representation of the first derivative is the slope of the tangent at a given point. And second-order derivative represents the slope changes over the independent variable in the graph.

Mathematically, if \(y = f(x)\). Then

\(\frac{{dy}}{{dx}} = {f^\prime }(x)\)

So, if \({f^\prime }(x)\) is differentiable, then differentiate \(\frac{{dy}}{{dx}}\) with respect to \(x\). We get the second-order derivative as,

\(\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}} = {f^{\prime \prime }}(x)\)

So, the second derivative of a function determines the local maximum or minimum inflection point values.

Similarly, higher-order derivatives can also be defined in the same way.

Example: \(\frac{{{d^3}y}}{{d{x^3}}}\) represents the third-order derivative and \(\frac{{{d^4}y}}{{d{x^4}}}\) represents the fourth-order derivative.

• If the value of the second-order derivative is positive, then the graph of a function is upwardly concave.
• If the value of the second-order derivative is negative, then the graph of a function is downwardly open.

Second-Order Derivatives of Functions in Parametric Form

To find the second derivative of the function in the parametric form, we apply the chain rule twice.

Suppose, we have \(x = x(t)\) and \(y = y(t)\), then its parametric form in second order:

The first derivative of a function \(y(t)\) with respect to \(x(t)\) in parametric form can be directly calculated as

\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

The second-order derivative is \(\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}\)

Note: It is wrong to write the above formula as \(\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{{{d^2}y}}{{d{t^2}}}}}{{\frac{{{d^2}x}}{{d{t^2}}}}}\).

Solved Examples

Q.1. Find the value of \(\frac{{dy}}{{dx}}\), if \(x = a{t^3} + 2{t^2},y = {t^2}\) at \(t = 0\).
Ans: Given: \(x = a{t^3} + 2{t^2},y = {t^2}\)

Now, differentiate \(x\) with respect to \(t\), then we get

\(\frac{{dx}}{{dt}} = 3a{t^2} + 4t\)

Similarly, differentiate \(y\) with respect to \(t\), then we get

\(\frac{{dy}}{{dt}} = 2t\)

We know that,

\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{2t}}{{3a{t^2} + 4}}\)

Now, when \(t=0\), then \(\frac{{dy}}{{dx}} = 0\)

\(\therefore \frac{{dy}}{{dx}} = 0\)

Q.2. Find the value of \(\frac{{dy}}{{dx}}\), if \(x = {e^t} + \sin t,y = {t^2}\) at \(t=0\).
Ans: Given: \(x = {e^t} + \sin t,y = {t^2}\)

Now, differentiate \(x\) with respect to \(t\), then

\(\frac{{dx}}{{dt}} = {e^t} + \cos t\)

Similarly, differentiate \(y\) with respect to \(t\), then

\(\frac{{dy}}{{dt}} = 2t\)

We know that, \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{2t}}{{{e^t} + \cos t}}\)

Now, when \(t=0\), then \(\frac{{dy}}{{dx}} = 0\)

\(\therefore \frac{{dy}}{{dx}} = 0\)

Q.3. Find the value of \(\frac{{dy}}{{dx}}\), if \(x = t\sin t,y = \cos t\) at \(t = \frac{\pi }{2}\).
Ans: Given: \(x = t\sin t,y = \cos t\) at \(t = \frac{\pi }{2}\)

Now, differentiating \(x\) with respect to \(t\), we get

\(\frac{{dx}}{{dt}} = t\cos t + \sin t\)    (Using product rule \(\frac{d}{{dx}}[f(x)g(x)] = f(x)\frac{d}{{dx}}[g(x)] + g(x)\frac{d}{{dx}}[f(x)]\))

Similarly, differentiating \(y\) with respect to \(t\), we get

\(\frac{{dy}}{{dt}} = – \sin t\)

We know that, \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{ – \sin t}}{{t\cos t + \sin t}}\)

When \(t = \frac{\pi }{2}\), we get,

\(\frac{{dy}}{{dx}} = \frac{{ – \sin \left( {\frac{\pi }{2}} \right)}}{{\frac{\pi }{2}\cos \left( {\frac{\pi }{2}} \right) + \sin \left( {\frac{\pi }{2}} \right)}}\)

\(\therefore \frac{{dy}}{{dx}} = – 1\)

Q.4. Find the value of \(\frac{{{d^2}y}}{{d{x^2}}}\) if \(y = {x^2} + 2x + 2\).
Ans: Given: \(y = {x^2} + 2x + 2\)

Differentiating with respect to \(x\), we get

\(\frac{{dy}}{{dx}} = \frac{{d\left( {{x^2} + 2x + 2} \right)}}{{dx}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{d\left( {{x^2}} \right)}}{{dx}} + \frac{{d(2x)}}{{dx}} + \frac{{d(2)}}{{dx}}\)

\( \Rightarrow \frac{{dy}}{{dx}} = 2x + 2 + 0\)

\( \Rightarrow \frac{{dy}}{{dx}} = 2x + 2\)

Again, differentiating with respect to \(x\), we get,

\(\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{d(2x + 2)}}{{dx}}\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{d(2x)}}{{dx}} + \frac{{d(2)}}{{dx}}\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 2 + 0\)

\(\therefore \frac{{{d^2}y}}{{d{x^2}}} = 2\)

Q.5. Find the value of \(\frac{{{d^2}y}}{{d{x^2}}}\) if \(y = {e^{\left( {{x^3}} \right)}} – 4{x^4}\).
Ans: Given: \(y = {e^{\left( {{x^3}} \right)}} – 4{x^4}\)

Now, differentiating with respect to \(x\), we get

\(\frac{{dy}}{{dx}} = {e^{\left( {{x^3}} \right)}} \times \frac{{dy}}{{dx}}\left( {{x^3}} \right) – 16{x^3}\)

\( \Rightarrow \frac{{dy}}{{dx}} = {e^{\left( {{x^3}} \right)}} \times 3{x^2} – 16{x^3}\)

Again, differentiating with respect to \(x\), we get,

Using product rule \(\frac{d}{{dx}}[f(x)g(x)] = f(x)\frac{d}{{dx}}[g(x)] + g(x)\frac{d}{{dx}}[f(x)]\), we get,

\(\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = {e^{\left( {{x^3}} \right)}} \times \frac{d}{{dx}}\left( {3{x^2}} \right) + 3{x^2} \times \frac{d}{{dx}}\left[ {{e^{\left( {{x^3}} \right)}}} \right] – 48{x^2}\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = {e^{\left( {{x^3}} \right)}} \times 3{x^2} \times 3{x^2} + {e^{\left( {{x^3}} \right)}} \times 6x – 48{x^2}\)

\(\therefore \frac{{{d^2}y}}{{d{x^2}}} = x{e^{\left( {{x^3}} \right)}} \times \left( {9{x^3} + 6} \right) – 48{x^2}\)

Q.6. Find \(\frac{{{d^2}y}}{{d{x^2}}}\), if \(y = {x^3}\).
Ans: Given: \(y = {x^3}\)

Now, let us first find the first-order derivative

Differentiate with respect to \(x\), then \(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{x^3}} \right) = 3{x^2}\)   [Using, \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n – 1}}\)]

Again, we will differentiate with respect to \(x\), to find the second derivative

Therefore, \(\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {3{x^2}} \right)\)

\(\therefore \frac{{{d^2}y}}{{d{x^2}}} = 6x\)

Summary

A parametric relation between \(x\) and \(y\) can be expressed as \(x = f(t)\) and \(y = g(t)\) is the parametric form representation with the parameter \(t\), such that \(y\) is defined as a function of \(x\). Note that each of the coordinates is expressed as another function of some parameter. We learned a method to find the derivative of the parametric function. Then, we discussed how a chain rule would be used to find the derivative of a parametric function with the help of standard differentiation formulas. Later, we discussed how to find second-order derivatives from the definition of first-order derivatives with the help of some examples.

Frequently Asked Questions (FAQs)

Q.1. How do you find the second derivative of a parametric function?
Ans: To find the second-order derivative of the function in the parametric form, we use the chain rule twice. Suppose we have \(x = x(t)\) and \(y = y(t)\). Hence the second-order derivative of a parametric function can be calculated as is
\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}\)

Q.2. What is the second derivative test used for?
Ans: Second derivative test used for determining:

• The local maxima or local minima, or points of inflection.
• The rate of change of the first derivative.
• Whether the slope of the tangent is increasing or decreasing.

Q.3. How do you differentiate parametric functions?
Ans: A relation between \(x\) and \(y\) can be expressed as \(x = f(t)\) and \(y = g(t)\). This is called the parametric representation with parameter \(t\), such that \(y\) is defined as a function of \(x\). So the first-order derivative is \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\), where \(\frac{{dx}}{{dt}} \ne 0\)

Q.4. What is a second-order derivative?
Ans: In general, the second-order derivative is defined as the derivative of the first-order derivative of the given function. The first-order derivative at a given point is the same as the slope of the tangent at that point. It is also defined as the instantaneous rate of change of a function at that point. Similarly, the second-order derivative gives us the idea of the shape of the graph of a given function.

Q.5. What is meant by parametric function?
Ans: Parametric functions are functions of several coordinates (\(2\) for the \(2\)-dimensional plane, \(3\) for \(3\)- dimensional plane, and so on), where each of the coordinates is expressed as another function of some parameter.
For example: For a function, \(y = f(x)\), the parametric functions are \(x = g(t),y = h(t)\).

Learn about Algebra of Derivatives of Functions Here

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