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

Oct 18, 2017

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

Explanation:

We seek $f ' \left(x\right)$, where:

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

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:

$f ' \left(x\right) = \left(\ln x + x\right) \left(\frac{d}{\mathrm{dx}} \left(\sin x + {e}^{x}\right)\right) + \left(\frac{d}{\mathrm{dx}} \left(\ln x + x\right)\right) \left(\sin x + {e}^{x}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\ln x + x\right) \left(\cos x + {e}^{x}\right) + \left(\frac{1}{x} + 1\right) \left(\sin x + {e}^{x}\right)$