How do you differentiate #f(x)=(sinx+lnx)(x^2-3e^x)# using the product rule?

1 Answer
Jun 20, 2017

Answer:

# f'(x) = (sinx+lnx)(2x-3e^x) + (cosx+1/x)(x^2-3e^x) #

Explanation:

We have:

# f(x) = (sinx+lnx)(x^2-3e^x) #

We will apply the Product Rule for Differentiation:

# d/dx(uv)=u(dv)/dx+(du)/dxv #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

# d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw#

So with # f(x) = (sinx+lnx)(x^2-3e^x) #;

# { ("Let", u = sinx+lnx, => , (du)/dx = cosx+1/x), ("And" ,v = x^2-3e^x, =>, (dv)/dx = 2x-3e^x ) :}#

Then # d/dx(uv)=u(dv)/dx + (du)/dxv # gives us:

# f'(x) = (sinx+lnx)(2x-3e^x) + (cosx+1/x)(x^2-3e^x) #