i have another math question, can someone please help?

Question

g(x)=5x^5+2x^1ln(x)...they want us to find g`(e), which is finding derivative of g(x) and then plugging in e.
.. i got stuck after 25x^4 =(

Answer 1

g(x) = 5x^5 + 2x lin(x)


now for x^ln(x)

let y = x^ln x
ln y = (ln x)^2
differentiate both sides

y dy/dx = 2 (ln x) 1/x
dy/dx = 2(ln x)/xy = 2(ln x)/x^(1+ln x)

so g' = 25x^4+2ln x/x^(1+ln x)

put x = e
g'(e) = 25e^4+2 /e^(1+1) = 25e^4+2/e^2

Answer 2

Well dear;
f(x) = 5x^5+2x^1ln(x)

Part a;
if y = 5x^5
Differentiate 5x^5 with respect to x.
Apply the "power rule" to x^5 (the derivative of xn=nxn-1) giving 25x^4

Part b;
you should use a trick in here
if y = 2x^ln(x)
so we have ;
ln y = (ln x)^2
now differentiate of each sides;
as you know derivative (lnx) = 1/x

y' = 2* (ln x) *1/x
y' = 2(ln x) / xy = 2(ln x) / x^(1+ln x)

so;
g'(x) = 25x^4+[2ln x / x^(1+ln x)]
now if x= e so fill it into the function;
The Total Result is ;

g'(e) = (25e^4)+(2 /e^(1+1)) = (25(2.17)^4)+ (2 /(2.17)^2 )

Good Luck.

Answer 3

d g/dx = 25 x^4+2x. 1/x+ln(x).2
g(e) which is = 25x 2.1714^4+2+2 since ln(e) = 1 and x and x cancells of previously.

Answer 4

g(x)=5x^5+2x^ln(x)

I thought it was easy, but I first have to read again some theorem about have make differentiation, Sorry but I will be back later

Answer 5

g(x)=5x^5+2x^1logx
derivative of 2x^1logx
=2(x*1/x+1*logx) using product rule
=2(1+logx)

g'(e)=25e^4+2(1+loge) loge=1
g'(e)=25e^4+4

Answer 6

that solution for differential of 2x^ln(x) is incorrect, you can see this by integrating