Question

find y'
x^(sqrt x)

Answer 1

\[\large y= x^{\sqrt {x}}\]\[\large \ln y=\sqrt {x}\large \ln x\]take derivative

Answer 2

y'= lnx/2sqrtx + sqrt x * 1/x)

Answer 3

stuck in this part

Answer 4

forget the y
y'= lnx/2sqrtx + (sqrt x * 1/x)y
y=x^sqrtx

Answer 5

y = x^(x^(1/2))

take the ln of both sides

ln(y) = x^(1/2)ln(x)

Answer 6

Solve it according to mukushla..

Answer 7

\[\frac{1}{y}dy = (\sqrt{x} \frac{1}{x} + \ln(x) \cdot \frac{1}{2 \sqrt{x}} )dx\]

Answer 8

ln(y) = x^(1/2)ln(x)

we have

\[y' * \frac{1}{y} = \frac{\ln(x)}{2x^{-1/2}} + \frac{x^{1/2}}{x}\]

Answer 9

In denominator power will be positive @Australopithecus

Answer 10

yeah thanks for noticing :)

Answer 11

So multiply by y both the sides..

Answer 12

\[y' = (\frac{\ln(x)}{2x^{1/2}} + \frac{x^{1/2}}{x})y\]

Answer 13

we know what

y=x^(x^(1/2))

so thus,

\[y' = (\frac{\ln(x)}{2x^{1/2}} + \frac{x^{1/2}}{x})x^{x^{1/2}}\]

Answer 14

\[\frac{dy}{dx} = (\sqrt{x} \times \frac{1}{x} + \ln(x) \cdot \frac{1}{2 \sqrt{x}} ) \times x^{\sqrt{x}}\]

Answer 15

i got it

Answer 16

x^1/2-x-1

Answer 17

Well done then..

Try to do more problems like this..

Answer 18

my final answ is 1/2sqrtx(ln x+2) x^sqrtx