Asked by anonymous
                find the second derivative of each of the following functions:
a) y=e^x + sin^2 (x)
b) y = 2√(x) + ln (x^3)
            
        a) y=e^x + sin^2 (x)
b) y = 2√(x) + ln (x^3)
Answers
                    Answered by
            Reiny
            
    To do these, you must have in your repertoire the ability to quickly differentiate
e^x, ln(something), sinx, cosx, and √(something) etc
I will do b) , you try the first one
y = 2√(x) + ln (x^3)
y = 2 x^(1/2) + ln(x^3)
y' = (1/2)(2) x^(-1/2) + 3x^2/x^3
= x^(-1/2) + 3x^-1
y '' = (-1/2)x^(-3/2) - 3x^-2
= -1/(2√x)^3 - 3/x^2 or variations of that after rationalizing the denominator
I got it simplified as
-1/(2x√x) - 3/x^2
    
e^x, ln(something), sinx, cosx, and √(something) etc
I will do b) , you try the first one
y = 2√(x) + ln (x^3)
y = 2 x^(1/2) + ln(x^3)
y' = (1/2)(2) x^(-1/2) + 3x^2/x^3
= x^(-1/2) + 3x^-1
y '' = (-1/2)x^(-3/2) - 3x^-2
= -1/(2√x)^3 - 3/x^2 or variations of that after rationalizing the denominator
I got it simplified as
-1/(2x√x) - 3/x^2
                    Answered by
            anonymous
            
    for a) i got y' = e^x cos(2x) * 2
for b) i got a different answer, i got y' = (√(x) - 6)) / (2x^2)
i use the symbolab to check my answer and i got it correct?
    
for b) i got a different answer, i got y' = (√(x) - 6)) / (2x^2)
i use the symbolab to check my answer and i got it correct?
                    Answered by
            Reiny
            
    when simplified, it's the same thing
look at my first term:
-1/(2x√x)
= -1/(2x√x) * √x/√x
= -√x/(2x^2) <---- same as the first term of your answer
and of course my - 3/x^2 is the same as your - 6/(2x^2)
    
look at my first term:
-1/(2x√x)
= -1/(2x√x) * √x/√x
= -√x/(2x^2) <---- same as the first term of your answer
and of course my - 3/x^2 is the same as your - 6/(2x^2)
                    Answered by
            anonymous
            
    is my answer for a is correct?
    
                    Answered by
            Reiny
            
    a)
y=e^x + sin^2 (x)
y ' = e^x + 2sinx cosx
= e^x + sin (2x)
y'' = e^x + 2cos(2x) , you had them multiplies, could have been just a typo
    
y=e^x + sin^2 (x)
y ' = e^x + 2sinx cosx
= e^x + sin (2x)
y'' = e^x + 2cos(2x) , you had them multiplies, could have been just a typo
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.