Derivative of a sinusoidal function. Need some Expertise

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Homework Help Overview

The discussion revolves around finding the derivative of the function h(x) = 3e^sin(x + 2). Participants are exploring the application of the chain rule and the properties of exponential functions in the context of calculus.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the chain rule and the derivative of exponential functions, questioning the role of the natural logarithm in the differentiation process. There is also a consideration of the distinction between different forms of the function and their derivatives.

Discussion Status

Some participants have confirmed their derivative calculations and are seeking further clarification on the role of constants and logarithms in the differentiation process. There is an acknowledgment of the correctness of the initial derivative, and some participants express understanding of the concepts discussed.

Contextual Notes

Participants are navigating potential confusion regarding the interpretation of constants in the function and how they affect the differentiation process. There is also mention of a modified function, prompting further exploration of the derivative in that context.

chubbyorphan
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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!:confused:
 
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chubbyorphan said:
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!:confused:

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
 
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?
 
: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:eek:

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))
 

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