Derivative of a sinusoidal function. Need some Expertise

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SUMMARY

The derivative of the function h(x) = 3e^sin(x + 2) is correctly calculated as h’(x) = (3e^sin(x + 2))(cos(x + 2)). The discussion clarifies the application of the chain rule and the derivative formula for exponential functions, specifically f(x) = a^g(x), where a is a positive constant. The confusion regarding the natural logarithm (ln) is addressed, confirming that while ln(e) = 1, it does not alter the derivative's outcome. The participants also explore a modified function, h(x) = 3e^sin(5x + 2), leading to the derivative h’(x) = (15e^sin(5x + 2))(cos(5x + 2).

PREREQUISITES
  • Understanding of the chain rule in calculus
  • Knowledge of derivatives of exponential functions
  • Familiarity with the natural logarithm and its properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the chain rule in depth, focusing on its applications in calculus
  • Learn about the properties of logarithmic functions, particularly natural logarithms
  • Explore the product rule and its implications for derivatives of products of functions
  • Practice finding derivatives of more complex exponential functions
USEFUL FOR

Students studying calculus, particularly those preparing for exams involving derivatives of exponential functions, and educators seeking to clarify concepts related to the chain rule and logarithmic properties.

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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