Question

find the derivative of y=sin sqrt(1+x^2)

Answer 1

do it using chain rule....

Answer 2

i did but i got stuck at \[y'=(\sin)(1/2)(1+x ^{2})^{-1/2}+(1+x ^{2})^{1/2}(\cos)\]

Answer 3

hmmm addition sign???
see wht would be derivative of
\[\sin(x^2 + 1)\]
just asking??????

Answer 4

you use the product rule

Answer 5

and hey ur question is
\[\sin \sqrt{1 + x^2} \]

Answer 6

product rule is used in questions like
1) \[x^2 e^x\]
2)\[\sin(x) \tan(x)\]

Answer 7

yea so i got \[y'=(\sin)(1/2)(1+x^2)^{-1/2}(2x)+(1+x^2)^{1/2}(\cos)\]

Answer 8

then if im wrong how would you find the derivative of that

Answer 9

nah nah ...u don't use product rule here....
chain rule.....
like u see
derivative of sin(x^2 + 1) would be
\[2x \cos(x^2 + 1)\]

Answer 10

well the book has a different answer

Answer 11

hey its not the answer for the question u asked....

Answer 12

the question u askd the answer is
\[\frac{ x \cos \sqrt{1 + x^2} }{ \sqrt{1 + x^2} }\]

Answer 13

okay yea that right but how did u get that

Answer 14

for the question u askd
put
u = sqr rt 1 + x^2
so y = sin u
now \[\frac{ dy }{ dx } = \frac{ dy }{ du } \times \frac{ du }{ dx }\]

Answer 15

so how would you set that up

Answer 16

if \[u = \sqrt{1 + x^2}\]
then wht would be
\[\frac{ du }{ dx }\] ????????

Answer 17

never mind i got it thank you

Answer 18

u r most welcome....