Question

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Answer 1

Answer:

$\frac{df(x)}{dx}=2(1+lnx)$

Step-by-step explanation:

From our knowledge of derivatives we use the product rule approach, the product rule is stated below

For f(x)=u(x)v(x) the derivative is express as

$\frac{df(x)}{dx}=u(x)\frac{dv(x)}{dx}+v(x)\frac{du(x)}{dx}$

hence from the question f(x)=2xlnx,

we can assign u(x)=2x and v(x)=lnx

also the derivative of u(x)=2x is simply

$\frac{du(x)}{dx}=2$

and the derivative of v(x)=lnx can be express using the general log derivative i.e

$y=lnx\\\frac{dy}{dx}=\frac{1}{x}$

hence we have $\frac{dv(x)}{dx}=\frac{1}{x}$

if we substitute value into the general product expression stated earlier we arrive at

$\frac{df(x)}{dx}=2x\frac{1}{x}+lnx*2\\\frac{df(x)}{dx}=2+2lnx\\\frac{df(x)}{dx}=2(1+lnx)$