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

Jun 7, 2016

$f ' \left(x\right) = {x}^{2} {e}^{x} \left(x \ln x + 3 \ln x + 1\right)$

Explanation:

The product rule for three functions basically states that the derivative of the product of three functions equals the derivative of one function multiplied by the other two functions, added to the the other two incarnations of this where the other two functions are differentiated in the other function's place.

Mathematically, this can be written as

${\left(f g h\right)}^{'} = {f}^{'} g h + f {g}^{'} h + f g {h}^{'}$

So, for $f \left(x\right) = \left({x}^{3}\right) \left(\ln x\right) \left({e}^{x}\right)$, we see that

$f ' \left(x\right) = {\left({x}^{3}\right)}^{'} \left(\ln x\right) \left({e}^{x}\right) + \left({x}^{3}\right) {\left(\ln x\right)}^{'} \left({e}^{x}\right) + \left({x}^{3}\right) \left(\ln x\right) {\left({e}^{x}\right)}^{'}$

Note that:

$\left\{\begin{matrix}{\left({x}^{3}\right)}^{'} = 3 {x}^{2} \\ {\left(\ln x\right)}^{'} = \frac{1}{x} \\ {\left({e}^{x}\right)}^{'} = {e}^{x}\end{matrix}\right.$

Therefore

$f ' \left(x\right) = 3 {x}^{2} \left(\ln x\right) \left({e}^{x}\right) + {x}^{3} \left(\frac{1}{x}\right) \left({e}^{x}\right) + {x}^{3} \left(\ln x\right) \left({e}^{x}\right)$

$= 3 {x}^{2} \ln x \left({e}^{x}\right) + {x}^{2} {e}^{x} + {x}^{3} \ln x \left({e}^{x}\right)$

$= {x}^{2} {e}^{x} \left(x \ln x + 3 \ln x + 1\right)$