Question

y = (3+sin(x))^x
Find the derivative using implicit differientation

Answer 1

So for this one we take ln of both sides, so
ln(y) = ln(3+sin(x))^x
using log properties to rewrite
ln(y) = (x)(ln(3+sin(x))
ln(y) = (1/y)(y')

Answer 2

So on the right side of the eqn we will find the derivative using some power and (chain?) rule and multiply by y to solve for y'

Answer 3

(1/y)(y') = (x)(ln(3+sin(x))
hmm... what would i do from here

Answer 4

how did you go from
ln(y) = (x)(ln(3+sin(x))
to
ln(y) = (1/y)(y')

Answer 5

no i mean that ln(y) equals (1/y)(y') so I set (1/y)(y') equal to (x)(ln(3+sin(x))
and then I'm just confused on the right side.
like what do i do about (x)(ln(3+sin(x)) from here?

Answer 6

oh you applied the derivative to ln(y)?

Answer 7

yeah derivative of ln(y) = 1/y but you have to multiply by y'
Implicit differientation

Answer 8

yes the derivative of ln(y) is (y')/y

what is the derivative of (x)(ln(3+sin(x)) ?

by "derivative" I mean "derivative with respect to x"

Answer 9

it might help to think of it like this
f(x) = x
g(x) = 3+sin(x)

then
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
by the product rule

Answer 10

sorry I meant to say g(x) = ln(3+sin(x))

Answer 11

we use product rule
so x times derivative of (ln(3+sin(x)) which is 1/(3+sin(x) but chain rule so we multiply by the derivative of that which is cos(x)
So x)(cosx/3+sin(x)) + ln(3+sin(x)) *1

Answer 12

hmm... i may have messed up somewhere but how do we get the derivative of ln(3+sin(x))?

Answer 13

You should get \(\large \frac{x\cos(x)}{3+\sin(x)}+\ln(3+\sin(x))\)

Answer 14

cool, i got that

Answer 15

By this rule
\(\large f(g(x)) = \ln(g(x))\)

\(\large f'(g(x)) = \frac{1}{g(x)}*\frac{d}{dx}[g(x)]\)

\(\large f'(g(x)) = \frac{g'(x)}{g(x)}\)

Answer 16

If g(x) = y, then,

By this rule
\(\large f(y) = \ln(y)\)

\(\large f'(y) = \frac{1}{g(x)}*\frac{d}{dx}[y]\)

\(\large f'(y) = \frac{y'}{y}\)

which is what we got earlier

Answer 17

hmm yeah
so to solve for y'
y' = \(\color{#0cbb34}{\text{Originally Posted by}}\) jimthompson5910
You should get \(\large \frac{x\cos(x)}{3+\sin(x)}+\ln(3+\sin(x))\)
\(\color{#0cbb34}{\text{End of Quote}}\)
times what y equals which is (3+sin(x))^x

Answer 18

basically y times the derivative of the right which we found earlier?

Answer 19

Yes, so far we have
\(\large \frac{y'}{y} = \frac{x\cos(x)}{3+\sin(x)}+\ln(3+\sin(x))\)
which solves to
\(\large y' = y\left(\frac{x\cos(x)}{3+\sin(x)}+\ln(3+\sin(x))\right)\)
then the last step is to plug in the original equation (what y is originally equal to). Optionally you can simplify things a bit afterward.

Answer 20

ion know

Answer 21

alrighty thx. I understand it now tomorrow i have a quiz and I might get a problem similar to this so I'll see if I can do that right, ty once again Jimthompson : )

Answer 22

No problem