Finding the derivatives of functions

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SUMMARY

This discussion focuses on finding the derivatives of three specific functions using calculus techniques. The first function, y = 2x / ex, was initially miscalculated but corrected to apply the quotient rule accurately. The second function, f(x) = 2x ln(x2 + 5), was confirmed as correct. The third function, g(x) = ln x / (ex2 + 2), was also validated. The participants emphasized the importance of proper application of differentiation rules, particularly the quotient rule.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly differentiation.
  • Familiarity with the quotient rule for derivatives.
  • Knowledge of logarithmic differentiation and its applications.
  • Proficiency in manipulating exponential functions and their derivatives.
NEXT STEPS
  • Study the application of the quotient rule in detail.
  • Explore logarithmic differentiation techniques for complex functions.
  • Practice finding derivatives of exponential functions using various rules.
  • Review common mistakes in differentiation and how to avoid them.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, as well as educators looking to reinforce differentiation techniques.

lamerali
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Hi, I'm working with finding the derivatives of functions, which I'm not very comfortable with; if someone could please check my answers to the following questions i would be VERY grateful! Thank you! :)

find the derivative of the following function:

Question 1:

y = [tex]\frac{ 2^{x} }{ e^{x} }[/tex]

My Answer

y1 = [tex]\frac{ e^{x} . ln2 . 2^{x} + 2^{x} . e^{x} }{ e^{x}^{2} }[/tex]
= [tex]\frac{ 2^{x} (ln2 + 1) }{ e^{x} }[/tex]

Question 2:

f(x) = 2x ln(x[tex]^{2}[/tex] + 5)

My answer

f [tex]^{1}[/tex] (x) = 2ln(x[tex]^{2}[/tex] + 5) + (2x) . [tex]\frac{1}{x^{2} + 5}[/tex] . (2x)

= 2 ln(x[tex]^{2}[/tex] + 5) + [tex]\frac{4x^{2}}{x^{2} + 5}[/tex]

Question 3:

g(x) = [tex]\frac{ln x}{e^{x}^{2} + 2}[/tex]

My answer:

g[tex]^{1}[/tex](x) = [tex]\frac{(e^{x}^{2} + 2) . (1/x) - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

= [tex]\frac{\frac{e^{x}^{2} + 2}{x} - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

for the last two questions I'm not sure if i simplified enough...if anyone could guide me in the right direction where needed i'd really appreciate it! thanks in advance!
 
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lamerali said:
Hi, I'm working with finding the derivatives of functions, which I'm not very comfortable with; if someone could please check my answers to the following questions i would be VERY grateful! Thank you! :)

find the derivative of the following function:

Question 1:

y = [tex]\frac{ 2^{x} }{ e^{x} }[/tex]

My Answer

y1 = [tex]\frac{ e^{x} . ln2 . 2^{x} + 2^{x} . e^{x} }{ e^{x}^{2} }[/tex]
Almost right you need "-" , not "+" in the numerator- quotient rule: (u/v)'= (u'v- uv')/v^2.

= [tex]\frac{ 2^{x} (ln2 + 1) }{ e^{x} }[/tex]

Question 2:

f(x) = 2x ln(x[tex]^{2}[/tex] + 5)

My answer

f [tex]^{1}[/tex] (x) = 2ln(x[tex]^{2}[/tex] + 5) + (2x) . [tex]\frac{1}{x^{2} + 5}[/tex] . (2x)

= 2 ln(x[tex]^{2}[/tex] + 5) + [tex]\frac{4x^{2}}{x^{2} + 5}[/tex]
Yes, that looks good.

Question 3:

g(x) = [tex]\frac{ln x}{e^{x}^{2} + 2}[/tex]

My answer:

g[tex]^{1}[/tex](x) = [tex]\frac{(e^{x}^{2} + 2) . (1/x) - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

= [tex]\frac{\frac{e^{x}^{2} + 2}{x} - lnx . 2xe^{x}^{2}}{(e^{x}^{2} + 2)^{2}}[/tex]

for the last two questions I'm not sure if i simplified enough...if anyone could guide me in the right direction where needed i'd really appreciate it! thanks in advance!
The third problem also looks good to me.
 
Great! Thank you HallsofIvy! :D
 

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