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

Question

William Abdalla · Accepted Answer

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

The product rule states that if #f(x)=g(x)h(x)#, then

#f'(x)=g'(x)h(x)+g(x)h'(x)#

Here, we have

#g(x)=sinx+lnxcolor(white)(xxxx)=>color(white)(xxxx)g'(x)=cosx+1/x#

#h(x)=x-3e^xcolor(white)(xxxx)=>color(white)(xxxx)h'(x)=1-3e^x#

Consequently, we have

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

Evelyn Agard · Answer

undefined To differentiate $ f(x) = (sinx + lnx)(x - 3e^x) $ using the product rule, you would follow these steps:

1. Identify the two functions being multiplied: $ u(x) = sinx + lnx $ and $ v(x) = x - 3e^x $.
2. Apply the product rule formula: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Differentiate $ u(x) $ and $ v(x) $ separately.
   - $ u'(x) $ is the derivative of $ sinx + lnx $, which is $ cosx + \frac{1}{x} $.
   - $ v'(x) $ is the derivative of $ x - 3e^x $, which is $ 1 - 3e^x $.
4. Substitute the derivatives and the original functions into the product rule formula.
   - $ f'(x) = (cosx + \frac{1}{x})(x - 3e^x) + (sinx + lnx)(1 - 3e^x) $.
5. Simplify the expression if necessary.