Question

need to differentiate twice here, y''. [d^2y/dx^2]

Answer 1

this is the question:

Answer 2

Now whats with double differentiation ?is it your assignment

Answer 3

\[siny - e^x=5x^6\]

Answer 4

can u pls show the answer as d^2y/dx^2 = aanswer.... nd its kinda my assignment, im just doing practice questions and need some help with double differentiation.

Answer 5

Now when I differentiate

\[cosy \frac{dy}{dx} - e^x = 30x^5\]

Answer 6

Yea Sure

Answer 7

Then I double Differentiate
\[cosy \frac{d^2y}{dx^2} - siny \frac{dy}{dx} -e^x = 150x^4\]

Answer 8

maybe solve for
\[y'\] and then do it again would be easier. just a thought

Answer 9

First of all differentiate both sides once, you will get
cos y y'-e^x=30x^5
Now differentiate it again

Answer 10

Tell me what did u get?

Answer 11

\[\Large{ \frac{d^2y}{dx^2} = \frac{150x^4 + e^x + siny \frac{dy}{dx} }{cosx}}\]

Answer 12

\[\cos(y)y'-e^x=30x^5\]
\[y'=\frac{30x^5+e^x}{\cos(y)}\]

Answer 13

Thats what I get but wait for satellite

Answer 14

no your way may be easier. but you have a
\[cos(x)\] should be
\[\cos(y)\] and also you have a
\[\frac{dy}{dx}\] in your answer. but if you solve for it first, then you can solve the whole thing

Answer 15

Now use Quotient rule to the result shown by Satellite

Answer 16

i would solve for
\[y'\] first, then take the derivative again, and in place of
\[y'\] you put
\[\frac{30x^5+e^x}{\cos(y)}\]

Answer 17

give me a second and i will write it out

Answer 18

ah typo

Answer 19

an ugly mess is what you get!

Answer 20

dy/dx should be square in my answer

Answer 21

and cosy below it not cosx

Answer 22

\[\Large{\frac{d^2y}{dx^2} = \frac{150x^4 + e^x +siny \times (\frac{dy}{dx})^2}{cosy}}\]

Answer 23

ok i will type it out, but i am not doing the algebra. first off i get
\[y'=\frac{30x^5+e^x}{\cos(y)}\]

Answer 24

Maybe let satellite confirm

Answer 25

then i take the derivative of that using quotient rule to get
\[\frac{\cos(y)(120x^4+e^x)+(30x^5+e^x)\sin(y)y'}{\cos^2(y)}\]

Answer 26

that 120 should be 150

Answer 27

then i put\[y'=\frac{30x^5+e^x}{\cos(y)}\] to get
\[\frac{\cos(y)(150x^4+e^x)+(30x^5+e^x)\sin(y)\frac{30x^5+e^x}{\cos(y)}}{\cos^2(y)}\]

Answer 28

the algebra is on you

Answer 29

alright, i managed to reach that point as well.... i can do the rest, thanks all of u so much. (was forgetting to substitute the dy/dx into the 2nd part... o.o)

Answer 30

good luck with algebra!

Answer 31

guys plz http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e382f680b8bf47d065f213e