Question

If g(x) + xsin g(x) = x^2 , find g ' (0)

Answer 1

\[g(x) + x \sin g(x) = x ^{2}\]
find \[g ' (0) \]

Answer 2

take the derivative and get
\[g'(x)+\sin(g(x))+x\cos(g(x))g'(x)=2x\] now replace x by 0 and solve for
\[g'(0)\]

Answer 3

actually i am confused. is it
\[g(x) + x \sin(x) g(x) = x ^{2}\] or
\[g(x) + x \sin( g(x)) = x ^{2}\]?

Answer 4

i am going to guess it is the first one, since i don't think you can do the second.

Answer 5

you get
\[g'(x)+\sin(x)g(x)+x\cos(x)g(x)+x\sin(x)g'(x)=2x\]now replace x by 0

Answer 6

you get
\[g'(0)=0\]

Answer 7

.so its 2(0) so its 0?

Answer 8

that is what i got, yes just 0

Answer 9

thank u very much. btw can i ask u a question on teh chain rule?

Answer 10

im confused when to use it and how?