Calculus : ∫ xˣdx = d/dx(xˣ) ?? – Is it a Trap?

While studying Calculus , Did you ever think of what would have happened to a function f(x) = $x^x$ when its integration and derivative were equal? Under what special conditions this would have taken place ? Did you ever think of the role of constant of integration in this particular problem? If it is neglected what would happen ?. This is avery good problem for advanced calculus enthusiasts. The concepts discussed here are very useful for JEE advanced and JEE Mains aspirants also .

Here we take a function, f(x)= $x^x$ and find its derivative and integral as per the calculus principles.. Here we demonstrate that it is not only important to use the formulae and applications in calculus but it is also of much importance to adopt the logical approach.

We discuss in depth the process of integration and differentiation . Then apply the principle of finding the constant of integration. We also discuss the common mistakes we have to avoid to get the correct answer to the problem. It is a lecture notes on concepts .

Calculus – Operations on xˣ

The basic operations in calculus are

1. Differentiation – Dividing
2. Integration – Addition

Differentiation of of xˣ ( Graphical Approach )

Before proceeding to mathematically solving $\frac{d}{\mathrm{dx}}(x^x)$, let us examine the how a graph of xˣ appears.

The plot of xˣ Vs. x

Just imagine the function f(x)= xˣ is divided into large number of parts such that the distance between the points of separation on x-axis is infinitesimally small , then the corresponding value of separation distance on y-axis is also observed from the plot. Then the plot of derivative of the above function can be given as,

The plot of d/dx(xˣ) Vs. x

By close observation of both plots what are the various points of importance do you find? Guess for a minute. Some of the major points of observations are:

1. The curve of $\frac{d}{\mathrm{dx}}(x^x)$ Vs.x cuts and crosses the x-axis at exactly same point of where the curve xˣ has the minimum value.
2. Till the curve The curve of $\frac{d}{\mathrm{dx}}(x^x)$ vs. x cuts and crosses the x-axis, the curve xˣ goes down up to its minimum value.
3. At x=1, the value of both curves on y-axis is same .
4. The curve xˣ has upward U shape.
5. The curve $\frac{d}{\mathrm{dx}}(x^x)$ is continuously increasing.

The above are some of the observations of from the plots. Once we understand the concept of calculus through graphical means , we can appreciate its importance in major scientific applications.

Kindly make a note of other various observations you find in the above curves and write down them and send us an email or contact us through contact page for any further clarifications. We answer your questions.

Analytical Solution of d/dx(xˣ)

The analytical solution to find the values $\frac{d}{\mathrm{dx}}(x^x)$ with the help of calculus is as below :

Let y=f(x)= xˣ, by taking natural logarithm on both sides , we get,

lny=xlnx, by differentiation on both sides, we get,

$\frac{d}{\mathrm{dx}}(\ln y)$ = $\frac{d}{\mathrm{dx}}(x\ln x)$,

But $\frac{d}{\mathrm{dx}}(\ln y)=\frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}$ , hence,

$\frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{d}{\mathrm{dx}}(x\ln x)$.

$\frac{\mathrm{dy}}{\mathrm{dx}}=y\frac{d}{\mathrm{dx}}(x\ln x)$ . By applying the product rule in differentiation, we get,

$\frac{d(x^x)}{\mathrm{dx}}=x^x[x\frac{d}{\mathrm{dx}}(\ln x)+\frac{d}{\mathrm{dx}}(x\mathrm{).}\ln x]$

$\frac{d(x^x)}{\mathrm{dx}}=x^x[x.\frac{1}{x}+1.\ln x]$

[Hint : The product rule is $\frac{d}{\mathrm{dx}}(\mathrm{uv})=u\frac{\mathrm{dv}}{\mathrm{dx}}+v\frac{\mathrm{du}}{\mathrm{dx}}$ where u and v are functions of x]

Finally we get, $\frac{d(x^x)}{\mathrm{dx}}=x^x[1+\ln x]$

We can check the value of Finally we get, $\frac{d(x^x)}{\mathrm{dx}}=x^x[1+\ln x]$ for x=1 and also compare the value from the graph given above.

The values of original function and its derivative will be same at x=1. ( Check it as Home work).

At x=1/e . the function will have minimum value and it is called as the stationary point in this particular case. regarding this point we discuss in the upcoming articles.

Integration of xˣ

The integration of $x^x$ in regular way is written as $\int x^x\mathrm{dx}$. It is highly stupendous task to find the value of $\int x^x\mathrm{dx}$ in calculus. It requires higher calculus knowledge to solve it.

The condition here given is that, $\int x^x\mathrm{dx}=\frac{d}{\mathrm{dx}}(x^x)$

From the above derivations of derivative of $x^x$, we get,

$\int x^x\mathrm{dx}=x^x(1+\ln x)$

Mistake to be avoided ( Important )

The big mistake likely to be committed in above problem is that from The condition here given is that, $\int x^x\mathrm{dx}=\frac{d}{\mathrm{dx}}(x^x)$, and then for $\int x^x\mathrm{dx}=x^x(1+\ln x)$ , it differentiated both sides and get the answer as,

$x^x=\frac{d}{\mathrm{dx}}(x^x(\ln x+1\mathrm{))}$

The above assumption is erroneous as the constant of integration is is a function of x as C=F(x) and is constant for a particular value of x not a universal constant.

Avoid this type of fatal errors .You land in getting wrong answers.

Main Problem – Difficult step? : To find Constant of integration C=F(x).

The main problem occurs when we try to solve equation $\int x^x\mathrm{dx}=x^x(1+\ln x)$ . It is not a straight forward integration in calculus. It requires advanced techniques and assumptions to solve using various calculus methods.

Let us define a variable ‘t’ and constant of integration C as,

$\int x^x\mathrm{dx}={\int }_0^x{t}^t\mathrm{dt}+C$

Here comes the role of integral constant C, from above,

$C=\int x^x\mathrm{dx}-{\int }_0^x{t}^t\mathrm{dt}$

But we know from the problem and above derivations that,

$\int x^x\mathrm{dx}=x^x(1+\ln x)$ , hence,

$C=x^x(\ln x+1)-{\int }_0^x{t}^t\mathrm{dt}$

From the above equation we see that, C = function of x and can be written as C=F(x). Here F(x)= $C=x^x(\ln x+1)-{\int }_0^x{t}^t\mathrm{dt}$

From above, the value of x can be found out as,

$x=F^{-1}(C)$where C is defined as $C=x^x(\ln x+1)-{\int }_0^x{t}^t\mathrm{dt}$and can be found out to be constant for a particular value of x.

Conclusion

The final answer for the when $\int x^x\mathrm{dx}=\frac{d}{\mathrm{dx}}(x^x)$, then,

$x=F^{-1}(C)$where C is defined as $C=x^x(\ln x+1)-{\int }_0^x{t}^t\mathrm{dt}$

It can also be proved that F(x) is continuous, strictly increasing and surjective. These can be approved in the subsequent lecture notes wherever occasion arises.

Home work

Kindly check if there is any positive value of x>0 for F(x) and also conclude by treating this as home work.

For further clarity on the problem , kindly see my video published in my YouTube Channel maths&science by iitian whose link is given here.

https://www.youtube.com/@mathsciencebyiitian?sub_confirmation=1

Above subject video can also be visited from the link here. The link for the above video is :

https://mathsnscience.com/resources/mind-blowing-calculus-problem/

Thus above one of the finest calculus problems is solved. Try to avoid neglecting the constant of integration while solving definite integral problems in calculus. Thus calculus is a subject which requires not only applications of rules it also involves a logical approach in calculus to avoid any errors.

The main trap is promptigng us to assume that integration and differentiation can simply be cancelled. The given equality is a special condition, not a standard identity. Also, the usual power rule cannot be applied to $x^x$, because the exponent is variable.

Best of luck to every body. We will meet with some other good question to strengthen our knowledge in calculus.