Question

Find dy/dx
y=xlog2(x)

Answer 1

Do you know the product rule? You have that y is a PRODUCT of two functions of x. The first function is just "x" the second is log(x).

Answer 2

I got here:
lny=lnx + ln log2(x)

Answer 3

You are using log with a base of 2, right?

Answer 4

yes

Answer 5

Ok, so rewrite that as
$$
\log_2(x) = {\ln(x)\over \ln(2)}
$$

Answer 6

Where ln(x) is the natural log of x

Answer 7

$$
\cfrac{d}{dx}x\log_2(x)=\cfrac{d}{dx}x{\ln(x)\over \ln(2)}=\cfrac{1}{\ln(2)}\cfrac{d}{dx}\left [x\ln(x)\right ]
$$
Now apply product rule

Answer 8

do you know the product rule?

Answer 9

no

Answer 10

You have y that is a product of two functions of x so you need to apply the product rule

$$
y=f_1(x)\times f_2(x)\\
\cfrac{d}{dx}y=f_2(x)\cfrac {d}{dx}f_1(x)+f_1(x)\cfrac {d}{dx}f_2(x)\\
$$
Where
$$
f_1(x)=x
$$
and
$$
f_2(x)=\log_2(x)
$$
So now you just need to find
$$
\cfrac{d}{dx}f_1(x)
$$
and
$$
\cfrac{d}{dx}f_2(x)
$$
Does that all make sense?

Answer 11

In words, the derivatives of y is log2(x) times the derivative of "x" PLUS "x" times the derivative of log2(x). That is the product rule.