 Hello everyone, welcome to Sendum Academy YouTube channel. So today we have got for you yet another question on differentiability. So let us first read the question and try to see how we can solve this question. So here the question says let f of x be a piecewise function e to the power negative 1 by x square sine of 1 by x when x is not equal to 0 and it is 0 when x is equal to We have to find the value of the derivative of this function at x equal to 0. Okay, so the first and the foremost thing is we need to first check is this function continuous because continuity is a necessary criteria for differentiability because of the option number d we have to perform this test. So let us do that. So what is the left hand limit of this function as x tends to 0. So left hand limit at x equal to 0. So for that we have to use the expression limit x tends to 0 minus e to the power negative 1 by x square sine of 1 by x. Now dear students, please note down that this particular function is a bounded function. Right, what's a bounded function? Bounded function means its value is restricted between minus 1 to 1 and this function e to the power minus 1 by x square as x tends to 0 minus becomes e to the power minus infinity which is tending to 0. So this fellow tends to 0. So overall the product of something tending to 0 into a bounded value is actually a 0. So the left hand limit is actually a 0. Let us find out similarly what would be our right hand limit at x tends to 0. So right hand limit would be limit x tending to 0 plus the same expression e to the power negative 1 by x square sine of 1 by x and again the same story repeats is you are going to get a bounded function multiplied to something which is tending to 0. So giving you the result as 0. And what about the value of the function at 0? So it's only given to us in the question that the value of the function at 0 is a 0. So f of 0 is going to be a 0. Right, now since the left hand limit and the right hand limit and the value of the function at 0 they match that means the function is continuous at x equal to 0. Okay, so this function is continuous at x equal to 0. So there is no issue with the continuity. So let's now move on to checking the left hand derivative and right hand derivative at x equal to 0. So let's find that out because the left hand derivative and the right hand derivative values if they are equal then only the function will be differentiable at x equal to 0 and the value itself will be your derivative of the function. So let's find it out. Now for finding the left hand derivative at x equal to 0 we are now going to use the first principles. So f dash 0 minus that will be limit of h tending to 0 plus f of 0 minus h minus f of 0 upon minus h. So what is going to be your f of 0 minus h? f of 0 minus h will be f of minus h and f of minus h will be e to the power minus 1 by h square sin of 1 by minus s. So minus will come out and f of 0 is only given to us as a 0 so no need to worry about that and your minus h comes down in the denominator. So this simplifies to limit s tending to 0 plus e to the power minus 1 by s square sin of 1 by h upon h. Now how to evaluate this limit? Let's try to see that. So for this we are going to split the entire limit into two parts. One is e to the power minus 1 by h square upon h and limit of limit of sin of 1 by h h tending to 0 plus. Now let us see this particular expression on the left side first. So let me call this fellow as L1 and let's call this as L2. Now L2 is something which is actually bounded. Please note L2 is something which is bounded. It is somewhere between minus 1 to 1. Okay so I do not know the value because this is an oscillating function sin 1 by x, oscillates very vigorously in the neighborhood of 0. So we don't know its exact value but what we know is that it's a bounded value. It's a bounded value. It's value lies between minus 1 to 1. Okay now about the limit L1. So let's find out the limit L1. So for the limit L1 we'll have to use our L'Hopital's rule. So what we can do is we can write this expression as limit sin to 0 plus 1 by h divided by e to the power 1 by h square. So mind you my dear students, this is actually infinity by infinity form. So you can actually apply L'Hopital rule on this. So you apply L'Hopital rule on this. You have to differentiate the numerator first and differentiate the denominator and then take the limit again. So derivative of the numerator will be negative 1 by x square. Derivative of the denominator will be e to the power 1 by s square into minus 2 by h cube. Right so this gives you the expression limit s tending to 0 plus h will come on the top. We'll have 2 e to the power 1 by h square. So this is tending to 0 divided by something which is very very large. Right so this quantity has to go to 0. Right so what about the entire limit? So the entire left hand derivative at x equal to 0 will be nothing but something tending to 0 into a bounded value. Right and this is going to become a 0 for sure. Right in a similar way in a similar way you can also find out the right hand derivative at x equal to 0. So right hand derivative at x equal to 0 will also be having the same fate and this is going to be a 0 value. Okay so left hand derivative and right hand derivative at x tends to 0 are both equal to 0 which means the derivative of the function at 0 is 0. So which option is correct? Let's check. So option number c, option number c is the right option for this particular question. I hope dear students you learn few concepts and especially the use of first principles to evaluate the derivative. Many of the students would actually like to use the direct method of derivative to get it but in that case your results will not be coming up as smoothly as you got in this case. Thank you so much for watching. Please do like, subscribe and comment.