Question

Differentiate y=(x+8)^sinx

the equation so it is easier to read in post. I will also post my steps and would like someone to tell me if I am correct. (many steps will take some time)

Answer 1

\[y=(x+8)^{sinx}\]

Answer 2

step 1:
\[y=(x+8)^{sinx}*ln(x+8)(sinx)\]

Answer 3

Woah, I don't understand that first step...

Answer 4

It looks like you tried to apply.... power rule... and exponential rule at the same time or something? Hmm

Answer 5

we were taught something like

\[U(X)^{v(x)} = U(X)^{V(X)}*(\ln(U(X))*V(X))\]

Answer 6

for that step 1

Answer 7

Hmm

Answer 8

Right. So first step would be take the natural log of both sides.

Answer 9

I have many other steps, but no need to go further if they are all wrong

Answer 10

so ln(y) = ln(x+8)^sinx

Answer 11

\[\large\rm y=u^v\qquad\to\qquad \ln y=\ln(u^v)\qquad\to\qquad \ln y=v \ln(u)\]Yah you can do logarithms to help you along.

Answer 12

\[\ln y = \ln(x+8)\sin(x)\]

Answer 13

So..

Answer 14

is that * sinx? or

Answer 15

Yes

Answer 16

so \[\frac{ 1 }{ y }y \prime = \frac{ 1 }{ x+8 } * sinx\]

Answer 17

You got to apply the full on product rule

Answer 18

ok but why would you do the product rule before doing what ever the Ln rule is called, where the derivative of ln is 1/x * d/dx of x

Answer 19

You have two functions on the right side of the equation you have a product of, so you did half of it. The deriv of the ln part times the sinx ...

Answer 20

ok I will post my results shortly

Answer 21

\[\ln(y) = \ln(x+8)(cosx) + (\frac{ 1 }{ x+8 }*(x+8)\prime)(sinx)\]

Answer 22

\[\frac{ 1 }{ y }y \prime = \ln(x+8)(cosx) + (\frac{ 1 }{ x+8 })*sinx\]

Answer 23

that would reduce to

\[\frac{ 1 }{ y }y \prime = \ln(x+8)(cosx)+(\frac{ sinx }{ cosx })\]

Answer 24

messed up on that last one

Answer 25

When you differentiate you are trying to find the derivative y prime, right? So all you need to do is multiply each side by y...

Answer 26

\[\frac{ 1 }{ y }y \prime = \ln(x+8)(cosx) + (\frac{ sinx }{ x+8 })\]

Answer 27

Keeping the function in terms of x

Answer 28

not sure what you mean

Answer 29

so ideally you end up having y'=...... noting that y=f(x) was already defined at the beginning of the problem

Answer 30

yes that is what I want lol

Answer 31

This identity gives a simple way to do the differential :P
\[a^b=(e^{\ln(a)})^b=e^{bln(a)}\]

Answer 32

so then \[e ^{sinx \ln(x+8)}\]

Answer 33

yup :) Then apply chain rule

Answer 34

to just the exponent? or do exponent rule and then chain?

Answer 35

So use the chain rule after using my identity. You've got to try and think of a good substitution to use :)

Answer 36

then would it be

e\[e ^{\frac{ sinx }{ x+8 }+(cosx)(\ln(x+8))}\]

Answer 37

Nope

Answer 38

\[y=(x+8)^{sinx}=e^{\sin(x)\ln(x+8)}\]

Answer 39

Use the substitution \[u=\sin(x)\ln(x+8)\] for the chain rule

Answer 40

then u = \[\frac{ sinx }{ x+8 } + (cosx)(\ln(x+8))\]

Answer 41

$$\frac{du}{dx}$$ is equal to that, yes. Not u.

Answer 42

i used product rule

Answer 43

I am sorry, but I am just completely lost at where we are.

Answer 44

I really do appreciate the help, I will go watch some youtube videos and then come back

Answer 45

We wish to find
\[\frac{dy}{dx}\].
But
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] by the chain rule.
Now, if we let
\[u=sin(x)ln(x+8)\]
then:
\[y=e^u\]
since
\[y=e^{sin(x)ln(x+8)}\]
So we can find
\[\frac{dy}{dx}\] by finding
\[\frac{dy}{du}\] and \[\frac{du}{dx}\]

Answer 46

Thank you for that information

Answer 47

:)