 Hello and welcome to the session. The question says, find the derivative of the following functions from the first principle, third part is sin x plus 1. Let's start with the solution and let us denote the given function by fx. Therefore, f at x plus h is equal to sin x plus h plus 1. Now let us find the first derivative of fx which is given by limit as h approaches to 0, f at x plus h minus fx upon h and it is denoted by f dash x where f is a real valued function. This is defined wherever the limit exists. Now let us substitute the values of fx plus h and fx we have sin x plus h plus 1 minus sin x plus 1 upon h. Now the formula of sin a minus sin b is 2 cos a plus b upon 2 into sin a minus b upon 2. So, here a is x plus h plus 1 and b is x plus 1. So, this can further be written as limit as h approaches to 0. Two times of cos on adding we have 2x plus h plus 2 upon 2 into sin a minus b is h upon 2 and in the denominator we have h. This is further equal to limit as h approaches to 0 x plus 1 plus h upon 2 into sin h upon 2 upon h or it can further be written as limit as h approaches to 0 cos x plus 1 plus h upon 2 taking this 2 in the denominator we have sin h upon 2 upon h upon 2. This is further equal to limit as h approaches to 0 cos x plus 1 plus h upon 2 into limit as h approaches to 0 sin h by 2 upon h by 2. Now h approaches to 0 implies h upon 2 also approaches to 0 and limit as h upon 2 approaches to 0 sin h upon 2 upon h upon 2 is equal to 1. Therefore, value of this limit is 1 and thus we further have this is a rough work limit as h approaches to 0 cos x plus 1 plus h upon 2 into 1. Now as h approaches to 0 this implies this term is 0 and thus further gives cos x plus 1. That is the derivative of the given function but the help of first derivative you find out and this is cos x plus 1. So, this completes the third part. Hope you have understood it. Take care and have a good day.