Susmita K.

y'=ddx[sin3x+cos4x]=3cos3x?4sin4x Explanation: We're going to use a few rules here to help us come up with our answer. The first rule is that when you're adding functions, the derivative of the sum is the same as the sum of the derivatives. We would write this mathematically as ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]. Quick note: the derivative of the function f(x) can be written as ddx[f(x)] or as f'(x). So, we know that ddx[sin3x+cos4x]=ddx[sin3x]+ddx[cos4x]. Ok, in order to do the derivatives ddx[sin3x] and ddx[cos4x], we're going to need to need a few more rules. ddx[sinx]=cosx. ddx[cosx]=?sinx. But how does that help us find ddx[sin3x]? To do ddx[sin3x] we have to use the chain rule. The chain rule says ddx[f(g(x))]=f'(g(x))?g'(x). Let's work through that with an example. Let f(x)=sinx and g(x)=3x, so f(g(x))=sin3x. Then f'(x)=cosx and g'(x)=3. By the chain rule, ddx[sin3x]=ddx[f(g(x))]=f'(g(x))?g'(x). So subbing in what we know gives ddx[sin3x]=cos(3x)?3, which we would write as ddx[sin3x]=3cos3x. Similarly, ddx[cos4x]=?sin(4x)?4=?4sin4x. So our final answer is y'=ddx[sin3x+cos4x]=3cos3x?4sin4x.

Have a thorough knowledge of your textbook Chapters. Don't Simply learn the Questions from any Guide. Try to form your own questions. Because in 10th you will get questions from textbook only not from your notes and all. So prepare in such a way that you can answer any question from text book. tudents says question is out of syllabus when question is asked from the chapter itself. The reason is students doesn't read text books.Put a habit of reading text books if you doesn't have. After thorough Preparation, collect previous question papers and try to solve the questions within 3 hours.In this way you score good marks. All the very Best to you.