Derivative of a sinusoidal function. Need some Expertise

[chubbyorphan]

Hey forum, thanks in advance to anyone who can explain this to me!

Homework Statement

Find the derivative of each of the following:
h(x) = 3e^sin(x + 2)

The Attempt at a Solution

My friend and I worked through this together and solved for:
h’(x) = (3e^sin(x + 2))(cos(x + 2))

Using my graphing calculator I have already confirmed this answer..

Homework Equations

What I need help with is just a little bit of light being shed on the concept..

My book says : the chain rule says the derivative of u^(-1) is (-1)u^(-2) u'
so I get that..
my book also says: the derivative of the function f(x)=a^g(x), where a is a positive constant, is given by f'(x)=a^g(x) lna g'(x)

so applying to the question above.. isn't 3e a positive constant..

so my question basically is.. when working this problem through..
is there a middle step where a 'lne' is established?

I realize lne = 1 anyway.. but I'm just wondering for the sake of making sure I thoroughly understand.

and if there is a middle step involving the natural logarithm(ln) why wouldn't it be ln3e
I realize its not because that would change the answer that I already confirmed but I'm just struggling to understand why not..
cuz if a mid step with ln is involved.. 3 * e is a positive constant I thought so I'd figure it would be ln3e :S

or another option is its lne * 3 which still equals 1..

anyway, I'm rambling now so if someone could please explain this to me I would really appreciate it!

 

[Ray Vickson]

Hey forum, thanks in advance to anyone who can explain this to me!

Homework Statement

Find the derivative of each of the following:
h(x) = 3e^sin(x + 2)

The Attempt at a Solution

My friend and I worked through this together and solved for:
h’(x) = (3e^sin(x + 2))(cos(x + 2))

Using my graphing calculator I have already confirmed this answer..

Homework Equations

What I need help with is just a little bit of light being shed on the concept..

My book says : the chain rule says the derivative of u^(-1) is (-1)u^(-2) u'
so I get that..
my book also says: the derivative of the function f(x)=a^g(x), where a is a positive constant, is given by f'(x)=a^g(x) lna g'(x)

so applying to the question above.. isn't 3e a positive constant..

so my question basically is.. when working this problem through..
is there a middle step where a 'lne' is established?

I realize lne = 1 anyway.. but I'm just wondering for the sake of making sure I thoroughly understand.

and if there is a middle step involving the natural logarithm(ln) why wouldn't it be ln3e
I realize its not because that would change the answer that I already confirmed but I'm just struggling to understand why not..
cuz if a mid step with ln is involved.. 3 * e is a positive constant I thought so I'd figure it would be ln3e :S

or another option is its lne * 3 which still equals 1..

anyway, I'm rambling now so if someone could please explain this to me I would really appreciate it!

The function is $3 \,e^{\sin(x+2)},$ with the '3' just being a constant 'outside' all the other computations. If you wanted to exponentiate 3e you would need to write $(3e)^{\sin(x+2)}.$

RGV

 

[╔(σ_σ)╝]

h(x) = 3*(e^sin(x + 2))

If f(x)= c*g(x) then f'(x)= c*g'(x) ( you can verify this by product rule).

In this equation c= 3 and g(x) = e^sin(x+2).

You are confusing H(x) = (3e)^sin(x+2) and h(x)= 3*e^sin(x+2). They are two different functions with two different derievatives.The answer you intially got is correct. Do you understand why?

 

[chubbyorphan]

:D ohhhh yes I do.. damn that seemed really obvious >.>
so we are using:

the derivative of the function f(x)=a^g(x), where 'a' is a positive constant, is given by f'(x)=a^g(x) lna g'(x)

and for the question h(x) = 3e^sin(x + 2)
lne is established, however, simplified out later because it's value is 1
Right?

Thanks to both of you for your help :D
who knows how many marks you just saved me on my upcoming test

also.. if we slightly altered the question to:
h(x) = 3e^sin(5x + 2)

would the answer be:
h’(x) = (15e^sin(5x + 2))(cos(5x + 2))