 Hello and welcome to the session. In this session, we discuss the following question which says, find the derivative with respect to x of the function 2x squared plus 3x plus 4 upon square root of x. Before we move on to the solution, let's discuss some general theorems on differentiation to be used in this question. First we have the differential coefficient of the product of a constant k and the function fx is equal to the product of the constant k and the differential coefficient of the function fx. That is, d by dx of k into fx is equal to k into d by dx of fx. Where this k is the constant. The next theorem, where the differential coefficient sum of say u and v is equal to the sum of the differential coefficients of the two functions. That is, d by dx of u plus v is equal to du by dx plus tv by dx. Where this u and v are the functions of x, differential coefficients exist. Now, this can also be generalized as d by dx of u plus v plus w plus so on is equal to du by dx plus tv by dx plus d w by dx plus and so on. So, you can say the differential coefficients of the sum of two or many functions is equal to the sum of the differential coefficients of these functions. Where again these u, v, w and so on are the functions of x whose differential coefficients exist. So, this is the key idea that we use for this question. Let's proceed with the solution now. We suppose if the given function be y, that is, we take y equal to s3x plus 4 and this whole upon square root of x. And we are supposed to find respect to x, that is, we are supposed to find dy by dx. Now, this function y can be written as 2x square upon square root of x plus 3x upon this means y is equal to 2 into x to the power 3 upon 2 x to the power 1 upon 2 x to the power minus 1 upon 2. We have got this function y. Now, we know that dy by dx of u plus v plus w and so on is equal to du by dx plus dv by dx plus d w by dx and so on. So, using this we get dy by dx is equal to dy by dx of x to the power 3 upon 2 plus dy by dx of 3 into x to the power 1 upon 2 dy dx of 4 into x to the power minus 1 upon. Now, the dy by dx of the constant k into fx is equal to k into dx of fx. So, using this we can differentiate each of these. So, this gives us dx is equal to 2 into dy by dx of x to the power 3 by 2 3 into dy by dx of x to the power 1 upon 2 plus dy by dx of x to the power 1 upon 2. dy by dx is equal to 3 upon 2 into x to the power 3 by 2 minus 1 which is 1 upon 2 plus 1 upon 2 minus 1 which is x to the power minus 1 upon 2 plus 4 into minus 1 upon 2 into x to the power minus 1 upon 2 minus 1 which is minus 3 upon. So, the dy by dx is equal to, that is 2 into x to the power 1 upon 2 plus 3 upon 2 into x to the power minus 1 upon 2. Now, here 2 times 4. So, we have minus 2 into x to the power minus 3 upon 2. So, we have dy by dx is equal to 3 into square root of x plus 3 upon 2 upon x to the power 3 upon 2 derivative of the function with respect to x. So, this is our final answer. This completes the session. Hope you have understood the solution of this question.