Question

find dy/dx of y=log(e^x+sinx)​

Answer 1

Answer:

It is your answer hope it helps you.

Answer 2

 Note : -  d(logx)/dx = 1/x  d()/dx =   d(sinx)/dx = cosx  d(u+v)/dx = [d(u)/dx] + [d(v)/dx]   If y = f(g(x))  Then dy/dx = f'(g(x))*g'(x)  For example if y = log(logx)  Then dy/dx = d[log(logx)]/dx*(d(logx)/dx)  = (1/logx)*[d(logx)/dx]  = (1/logx)*(1/x)    It's known as chain rule.  :-)