a differentiate problem?

Question

What is the value of dy/dx at x=e if y=xlogex. I don’t know which step should do first. Should I put the value of x into the function y first, or differentiate y first.

Answer 1

xlogex means x times the log base e of x which is written as x(ln x).

Start with the function y = x(ln x).

Use the product rule of differentiation: If y=(f)(g), then dy/dx = (f)(g') + (f')(g).

dy/dx = x(1/x) + 1(ln x)

dy/dx = 1 + ln x

The value of dy/dx at x = e is

dy/dx = 1 + ln e
dy/dx = 1 + 1
dy/dx = 2

Answer 2

whenever this type of problem comes u have to differentiate y first,
dy/dx = x*1/ex + log ex
= 1/e + log ex
now at x=e;
dy/dx = 1/e + log e^2
hope this is the answer.

Answer 3

differentiate y first, because if you put the value of e in first then you´ll just be differentiating a constant and the answer would always be 0, whatever the y(x) is

Answer 4

If you put in first the value of x you get a
number and its derivative is 0 .so this is wrong.
Derivate and then put in the given value
y´= log ex + 1/ex *e*x= log ex+1 ( Isupposed log is based "e")At x = e you get
y´(e) = loge^2+1 =3

Answer 5

You must differentiate first.

y = x log(base e)(x)
dy/dx = x (1/x) + log(x) = 1 + log(x).
When x = e, dy/dx = 1 + log(e)
= 1 + 1
= 2.

Answer 6

first differentiate,,using product rule,, (which is the derivative of first function of x into the second function,, and the derivative of he second fucntion of x into the first function)

remember dy/dx of loge

that is,,, dy/dx of x into logex + dy/dx of logex into x

that is 1*logex + 1/x * x (the x cancels here )

therefore we have,, logex +1

now dy/dx = logex + 1 substitute x = e

now we have logee which is 1 so 1+1 = 2

Answer 7

no opt to do it right now. Simplify earlier differentiating. y = ln( e^(-x) + x e^(-x) ) y = ln ( e^(-x) ( a million + x ) ) Separate making use of a log sources, y = ln(e^(-x)) + ln(a million + x) Simplifying making use of a log sources, y = (-x) ln(e) + ln(a million + x) ln(e) = a million, y = (-x)(a million) + ln(a million + x) y = -x + ln(a million + x) a lot a lot more convenient to distinguish this. dy/dx = -a million + a million/(a million + x) dy/dx = [ (-a million)(a million + x) + a million ] / [ a million + x ] dy/dx = [ -a million - x + a million ] / [ a million + x ] dy/dx = -x/(a million + x)

Answer 8

Differentiate first:

y=xloge(x)

d/dy = (d/dx x) · loge(x) + x· (d/dx loge(x))
d/dy = 1 · loge(x) + x· 1/x
d/dy = loge(x) + 1

at x = e
this is
loge(e) + 1 =
1 + 1 =
2
___

Hope this helps.

Answer 9

y = x log x
dy/dx = 1.log x + (1/x).x
dy/dx = log x + 1
Now at x = e:-
dy/dx = 1 + 1 = 2