Question

Find db/da for a^3b^2 = sin(a). Please help. :((

Answer 1

\(\large a^3b^2 = \sin(a)\)

Answer 2

\(a\) and \(b\) are variables

Answer 3

\(\large a^3b^2 = \sin(a)\)

differentiate both sides with.respect.to \(a\)

Answer 4

so would it be 3a^2b^2 = cos(a)?

Answer 5

you need to use product rule for left hand side

Answer 6

got it

Answer 7

\(\large a^3b^2 = \sin(a) \)

differentiate both sides with.respect.to \(a\)

\(\large a^3(b^2)' + (a^3)'b^2 = \cos(a) \)

Answer 8

(a^3)(2b)+(b^2)(3a^2) = cos (a) right?

Answer 9

not quite,
b is a function of a here - b is dependent variable, so by chain rule, u need to differentiate it again

Answer 10

\(\large a^3(b^2)' + (a^3)'b^2 = \cos(a)\)

\(\large a^3(2b) b' + (3a^2)b^2 = \cos(a)\)

Answer 11

solve \(\large b'\)

Answer 12

btw, \(\large b'\) is same as \(\large \frac{db}{da}\)

Answer 13

so would it be -a^3(2b)/(3a^2)b^2 +cos(a)

Answer 14

try again

Answer 15

I prefer to just gave up haha. I can't think anymore but okay..

Answer 16

\(\large a^3(2b) b' + (3a^2)b^2 = \cos(a)\)

\(\large a^3(2b) b' = \cos(a) - (3a^2)b^2\)

Answer 17

dont give up lol^ we're done wid calc :)
its just algebra, solve \(\large b'\)

Answer 18

i think i got it. waitt

Answer 19

okie

Answer 20

-cos(a)-3a^2b^2/ (a^32b)?

Answer 21

correct ! except for that negative sign..

Answer 22

\(\large a^3(2b) b' + (3a^2)b^2 = \cos(a)\)

\(\large a^3(2b) b' = \cos(a) - (3a^2)b^2\)

\(\large a^3(2b) b' + (3a^2)b^2 = \cos(a)\)


\(\large b' = \frac{\cos(a) - (3a^2)b^2}{a^3(2b)}\)

Answer 23

or may be put it like this :

\(\large b' = \frac{\cos(a) - 3a^2b^2}{2a^3b}\)

Answer 24

oh haha that was supposed to be an equal sign.

Answer 25

but thank you :DDD.

Answer 26

cool :) u wlc !