 Namaste, Myself, Mr. Pirajdar Bara Saheb, Assistant Professor, Department of Humanities and Sciences, Walshchand Institute of Technology, Soulapur. In this video lecture, we will discuss Laplace Transforms of Derivatives. Learning outcome, at the end of this session, students will be able to find Laplace Transforms of Derivatives. Let us start with statement of Laplace Transform of Derivatives and its proof. Statement is, if Laplace Transform of f of t equal to f of s, then Laplace Transform of f dash of t is equal to s into f of s minus f of 0. Now, proof of this statement, by definition of Laplace Transform, we have Laplace Transform of f of t equal to integration with limit 0 to infinity e raise to minus s t f of t dt. In this result, replace f of t equal to f dash of t on both the side, we get Laplace Transform of f dash of t is equal to integration with limit 0 to infinity e raise to minus s t f dash of t dt. Now, integrate right hand side integration using by parts, creating first function u as e raise to minus s t and second function v as f dash of t. So, we get Laplace Transform of f dash of t is equal to in bracket keeping e raise to minus s t as it is, into integration of f dash of t with respect to t is f of t, bracket close with limit 0 to infinity minus integration with limit 0 to infinity. Now, derivative of e raise to minus s t with respect to t is minus s into e raise to minus s t into integration of f dash of t is f of t dt. Now, substitute the limits in the right hand side, we get Laplace Transform of f dash of t is equal to in bracket when s equal to infinity, we get e raise to minus infinity into f of infinity minus when s equal to 0, we get e raise to 0 into f of 0 bracket close. In second term, when we take minus s is outside, we get minus minus plus s into integration with limit 0 to infinity e raise to minus s t f of t dt. As we know that e raise to minus infinity equal to 0 and e raise to 0 is 1. Therefore, Laplace Transform of f dash of t equal to minus f of 0 plus s into this definite integration is denoted by Laplace of f of t. Therefore, Laplace of f dash of t equal to s into, but Laplace of f of t is denoted by f of s minus f of 0 as it is. Similarly, we can prove that Laplace Transform of f double dash of t is equal to s square into f of s minus s into f of 0 minus f dash of 0. And Laplace Transform of f triple dash of t equal to s cube into f of s minus s square into f of 0 minus s into f dash of 0 minus f double dash of 0 and so on. When we observe the right hand side of these three result, we come to know that all terms in the right hand side are negative except first term. And in first term, degree of s is same as order of derivative of f of t. And in next successive term, degree of s goes on decreasing by 1 and in the last term, s is absent. By knowing this concept in general, if successive n minus 1th derivative of f of t is exist, then Laplace of nth derivative of f of t can be written as s raise to n into f of s minus s raise to n minus 1 into f of 0 minus s raise to n minus 2 into f dash of 0 minus and so on minus n minus 1th derivative of f of t at t tends to 0, where f of 0 f dash of 0 f double dash of 0 f triple dash of 0 etcetera are the values of f of t f dash t f double dash of t f triple dash of t etcetera at t tends to 0 respectively. Note that this property is highly used in solving ordinary differential equation with initial conditions. Now, working rule to solve the problems. Step 1, first find Laplace transform of given function f of t and denote it by f of s. Step 2, using Laplace transform of derivatives, we write Laplace of f dash of t equal to s into f of s minus f of 0 Laplace of f double dash of t equal to s square into f of s minus s into f of 0 minus f dash of 0 minus f double dash of 0 etcetera. Step 3, put the value of f of s f dash of 0 f of 0 f double dash of 0 etcetera. Now, let us pause the video for a while and write the answer to the question. Question is if Laplace transform of y of t equal to y of s, then write Laplace transform of d square y by d t square. Come back, I hope you have written answer to this question. Here I will going to explain the solution. Question is if Laplace of y of t equal to y of s, then write Laplace transform of d square y by d t square. We know that Laplace of d square y by d t square can be written as Laplace of y double dash of t. By Laplace transform of derivatives, we know that Laplace of f double dash of t equal to s square into f of s minus s into f of 0 minus f dash of 0. Similarly, here Laplace transform of d square y by d t square, which is Laplace transform of y double dash of t can be written as s square into y of s minus s into y of 0 minus y dash of 0. This is the required answer. Now, let us consider the example, example 1. If f of t equal to sin t upon t, then find Laplace transform of f dash of t. Here given function sin t upon t can be denoted by f of t and now we have to find it is a Laplace transform. First we find Laplace transform of sin t, which is equal to 1 upon s square plus 1 because Laplace of sin a t equal to e upon s square plus a square standard formula. Now, by effect of division by t, Laplace transform of f of t upon t is equal to integration of f of s with respectivities between the limit s to infinity. In this, put f of t sin t in the left hand side and its Laplace transform f of s in the right hand side. We get Laplace of sin t upon t equal to integration of 1 upon s square plus 1 into d s between the limit s to infinity. And integration of 1 upon s square plus 1 with respect to s is tan inverse of s, where s running from s to infinity. First put s equal to upper limit infinity, we get tan inverse of infinity minus s equal to lower limit s, we get tan inverse of s. But we know that tan inverse of infinity is pi by 2. Therefore, Laplace of sin t upon t equal to pi by 2 minus tan inverse of s, which is equal to we can write cot inverse of s. Now, this result is denoted by f of s. Now, we have to find value of f of 0, which means limit of f of t at t tends to 0, which is equal to limit of sin t upon t at t tends to 0. From the standard result of limit, we know that value of limit of sin t upon t as t tends to 0 is 1. Therefore, value of f of 0 is equal to 1. Now, by Laplace transform of derivatives, we have Laplace of f dash of t equal to s into f of s minus f of 0 denote this result by 1. Now, substitute the value of f of s and f of 0 already obtained in this right hand side of equation 1, we get Laplace transform of f dash of t equal to s into cot inverse of s minus 1. This is the required answer to the u 1 question. Now, let us consider another example, example 2. If f of t equal to e raise to minus 5 t into cos of 2 t, then find Laplace transform of f dash of t. Again, here given function f of t equal to e raise to minus 5 t into cos 2 t. First, we find Laplace transform of this f of t. As we know that Laplace transform of cos of 2 t, which is equal to s upon s square plus 2 square, that is equal to s upon s square plus 4. This is due to Laplace transform of cos at equal to s upon s square plus standard formula. Now, by force shifting theorem, we discussed earlier that is Laplace transform of e raise to minus e t into f of t equal to f of s plus a. Therefore, by this property we can write Laplace transform of e raise to minus 5 t into cos of 2 t is equal to in the Laplace transform of cos of 2 t, that is s upon s square plus 4. We have to replace every s equal to s plus 5. That is equal to s plus 5 upon s plus 5 bracket square plus 4. This result is denoted by f of s. Now, we have to find value of f of 0. f of 0 means, value of f of t at t tends to 0 that is equal to value of e raise to minus 5 t into cos 2 t at t equal to 0, which is equal to now put the value of t equal to 0. We get e raise to 0 into cos of 0 and as we know that cos 0 is 1 and e raise to 0 is 1. Therefore, f of 0 is equal to 1. Now, by Laplace transform of derivatives, we can write Laplace of f dash of t equal to s into f of s minus f of 0. Now, in this result put the value of f of s and f of 0 already we obtained. We get Laplace transform of f dash of t is equal to s into value of f of s is s plus 5 upon s plus 5 bracket square plus 4 minus value of f of 0 is 1. Therefore, 1 hence Laplace transform of f dash of t is equal to in numerator s in bracket s plus 5 upon s plus 5 bracket square plus 4 minus 1. This is the required answer to the given question. Now, to prepare this video lecture I referred these two books as the references. Thank you.