How do you differentiate #f(x)=(lnx+x)(sinx+e^x)# using the product rule?

1 Answer
Oct 18, 2017

Answer:

# f'(x) = (lnx+x)(cosx+e^x) + (1/x+1)(sinx+e^x)#

Explanation:

We seek #f'(x)#, where:

# f(x) = (lnx+x)(sinx+e^x) #

We will needs several standard derivatives:

# {: (ul("Function"), ul("Derivative"), ul("Notes")), (f(x), f'(x),), (x^n, nx^(n-1), n " constant (Power rule)"), (sinx, cosx, ), (e^x, e^x, e" base of Natural Logarithms"), (lnx, 1/x, ), (uv, uv'+u'v,"(Product rule)" ) :} #

So the applying the product rule we have:

# f'(x) = (lnx+x)(d/dx(sinx+e^x)) + (d/dx(lnx+x))(sinx+e^x)#
# \ \ \ \ \ \ \ \ = (lnx+x)(cosx+e^x) + (1/x+1)(sinx+e^x)#