Question

Can some one plz help me in this question
find the exact gradient of the tangent to the curve y=e^x+ln x at the point where x=1

I know y'= 1/x times e^x+lnx
And then sub 1
But i am not getting answer of 2e

Answer 1

times?

Answer 2

As in * multiply

Answer 3

\[y=e^x+\ln(x)\]
\[y'=e^x+\frac{1}{x}\]

Answer 4

X+lnx are both power

Answer 5

e^(x+lnx)

Answer 6

oooh it is
\[e^{x+\ln(x)}=e^xe^{\ln(x)}=xe^x\]

Answer 7

got it

Answer 8

Srry i am on my ipad

Answer 9

back to PEDMAS with you

Answer 10

your derivative is wrong

Answer 11

Oh hw so

Answer 12

yah its 1+1/x dummy!

Answer 13

first off \[\large e^{x+\ln(x)}=e^xe^{\ln(x)}=xe^x\] and to take the derivative of that, use the product rule
there is no \(\frac{1}{x}\) in it

Answer 14

e^x = e^x and e^lnx will be y'= 1/x e^lnx

Answer 15

^ or you can keep it as the exponential and do
d/dx ( e^(fx))= f'x * e^f(x)

d/dx (e^(x+lnx))=(1+1/x)*e^(x+lnx)

Answer 16

what ever you like best

Answer 17

The derivative of e^x is not x

Answer 18

whoa hold the phone
\[\frac{1}{x}e^{\ln(x)}=1\]

Answer 19

Hahah thank u

Answer 20

you are welcome ! :)

Answer 21

there is really no reason ever to write \(e^{\ln(x)}\) since it is just \(x\)

Answer 22

I know but In hsc our exam they cut off marks for that