How Do You Simplify the Derivative Expression of \( y = e^{7x+4} \)?

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Homework Help Overview

The discussion revolves around simplifying the derivative expression of the function \( y = e^{7x+4} \). Participants are exploring the application of limits and properties of exponential functions in the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the limit definition of the derivative and how to manipulate the expression involving \( e^{7x+4} \). There are attempts to apply properties of exponents to simplify the expression, and questions arise about the disappearance of certain terms in the limit process.

Discussion Status

The discussion is active with various approaches being considered. Some participants are questioning the steps taken to simplify the expression, while others are suggesting methods to handle the limit involving \( e^{7h} \). There is a recognition that understanding the derivative of \( e^x \) is crucial for progressing further.

Contextual Notes

There is some confusion regarding the interpretation of the final answer, with one participant asserting a numerical result rather than a functional derivative. This raises questions about whether the problem is asking for the derivative at a specific value of \( x \).

Chadlee88
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Could som1 please tell me what the next steps would be to be able to remove the h in the denomenator. :confused:

y = e^(7x+4)


Definition: lim f(x+h) - f(x)
h->0 h

lim (e^(7(x+h) + 4) - (e^(7x+4))
h->0 h

lim (e^(7x + 7h + 4)) - (e^(7x +4))
h->0 h
 
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How about using some of the properties of the exponential?
[tex]e^{7x+7h+4}= e^{7h}e^{7x+ 4}[/tex]
(yes, I could also have separated the "4" but it is the "h" that is important)
so
[tex]e^{7x+ 7h+ 4}- e^{7x+ 4}= e^{7x+4}(e^{7h}- 1)[/tex]
You will still have to deal with
[tex]\lim_{h\rightarrow 0}\frac{e^{7h}-1}{h}= 7\lim_{h\rightarrow 0}\frac{e^{7h}-1}{7h}[/tex]
and, taking k= 7h,
[tex]7\lim_{k\rightarrow 0}\frac{e^{k}-1}{k}[/tex]

but if you know how to deal with the derivative of ex you should be able to do that.
 
Last edited by a moderator:
i know that the final answer is 7 x .5 = 3.5 but i don't get how you got rid of e^(7x+4).

e^(7x+7h+4) - e^(7x+4) = e^(7x+4)(e^(7h) - 1) and then somehow
the e^(7x + 4) disappears and u get lim e^(7h-1)
h->0 h
 
Do you know how to derive the derivative of ex from the definition? If you can't, then you won't be able to solve this problem.
 
Chadlee88 said:
i know that the final answer is 7 x .5 = 3.5 but i don't get how you got rid of e^(7x+4).

e^(7x+7h+4) - e^(7x+4) = e^(7x+4)(e^(7h) - 1) and then somehow
the e^(7x + 4) disappears and u get lim e^(7h-1)
h->0 h
If that is the answer, then what is the question?

The derivative of e7x+4 is 7e7x+4! You don't "rid of" e7x+4, it's part of the answer. Since you assert that the answer is a number, 3.5, is it possible that the problem asks for the derivative at a given value of x?
 

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