Question

second order partial derivative of 2e - 2e^2y

Answer 1

y' = 2e - 2e^2y + 4e^2y =2e + 2e^2y
y" = 2e - 2e^2y + 4e^2y + 4e^2y + 8e^2y = 2e + 14e^2y

Answer 2

/??

Answer 3

@gerryliyana r u taking the derivative w.r.t. y??????/

Answer 4

nope.., thats wrong..,

Answer 5

according to me, if we are taking derivative w.r.t. y then
y'= 0-2(e^2y)*2 = -4e^2y
y''= -8e^2y

Answer 6

yes
y' = -4e^2y ln(e)
y" = -8*e^(2*y)*ln(e)^2
right??

Answer 7

@gerryliyana what does ln(e) signify in y' and ln(e)^2 signify in y''?????

Answer 8

hei @niksva relax..,hehe

y' = -4e^2y ln(e) = -4e^2y (1) = -4e^2y
y" = -8*e^(2*y)*ln(e)^2 =-8*e^(2*y) (1^2) =-8*e^(2*y)

right??
hint. ln(e) =1

Answer 9

Ok., @Rapi This is the complete solution:

\[u = 2e -2e ^{2y}\]
for first derivative:
\[\frac{ d }{ dy} (2e -2e ^{2y})\]
differentiate the sum term by term and facetor out constant
\[\frac{ du }{ dy } =2\left( \frac{ d }{ dy } (e) \right) - 2e ^{2}\left( \frac{ d }{ dy } (e^{2y}) \right)\]
derivative of e is zero
\[\frac{ du }{ dy } =2(0) - 2e ^{2}\left( \frac{ d }{ dy } (e^{2y}) \right)\]
using the chain rule, where s =2y
\[\frac{ d }{ dy } (e^{2y}) = e ^{s} \frac{ ds }{ dy }\]
and
\[\frac{ de ^{u} }{ u}=e ^{u}\]
\[-2e ^{2y}\left( 2\left( \frac{ d }{ dy } \left( y \right) \right) \right)\]

factor out contans, so u You will get the answer




for second derivative

\[\frac{ d ^{2}u }{ dy ^{2} } =\]

Answer 10

for second derivative
\[\frac{ d }{ dy } (-4e ^{2y}) \]
factor out constants
\[-4\left( \frac{ d }{ dy } (e ^{2y}) \right)\]
using the chain rule
\[\frac{ d }{ dy } (e ^{2y}) = e ^{S}\frac{ dS }{ dy } \]
where S =2y and
\[\frac{ de ^{S} }{ dS} = e ^{S}\]
becomes
\[-4\left( e ^{2y} \left( \frac{ d }{ dy }(2y) \right)\right)\]
Factor out constants
\[-4e ^{2y}\left( 2\left( \frac{ d }{ dy }(y) \right) \right)\]
How is the result of a derivative of y??

Good luck @Rapi!!