How Do You Apply the Chain Rule to Differentiate f(x) = x^5(4^(x^2))?

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Homework Help Overview

The problem involves finding the derivative of the function f(x) = x^5(4^(x^2)), which includes both polynomial and exponential components. Participants are discussing the application of the chain rule and product rule in this context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are exploring the correct interpretation of the function and discussing the differentiation process, particularly focusing on the product rule and the differentiation of exponential functions.

Discussion Status

The discussion is active with participants clarifying the function's form and sharing insights on differentiation techniques. Some guidance has been provided regarding the use of logarithms and the chain rule, though no consensus has been reached on the overall approach.

Contextual Notes

There is mention of potential confusion regarding the powers involved and the differentiation of the exponential term 4^(x^2). Participants are also acknowledging varying levels of familiarity with differentiation rules.

ninanana
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I'm so confused. I have to find the derivative of f(x) = x^5(4^(x^2)). All of the powers are messing me up. Any help would be much appreciated. Thanks!
 
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Is your function [tex]f(x) = x^5(4^{x^2})[/tex] or [tex]f(x) = x^{5(4^{x^2})}[/tex] ?
 
Last edited:
The first one, sorry.
 
Use product rule first, then you end up differentiating 4^(x^2).
A nice formula to know is d/dx ( a^(f(x)) ) = a^f(x) * ln(a) * f'(x), which comes from the chain rule.
 
I know I'm having some sort of stupid lapse right now, but the part I can't figure out is the 4^(x^2).
 
To differentiate [tex]y= 4^{x^2}[/tex], take the logarithm of both sides:
[tex]ln y= x^2 ln 4[/tex]
Now differentiate that, with respect to x.
[tex]\frac{1}{y}y'= 2x ln 4[/tex]
so
[tex]y'= 2x (ln 4)y= 2x 4^{x^2} ln 4[/tex]

Karlsen used the fact that the derivative of ax is ax ln a, but not everyone knows that!
 

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