find the second derivative of each of the following functions:
a) y=e^x + sin^2 (x)
b) y = 2√(x) + ln (x^3)
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
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?
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)
is my answer for a is correct?
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
To find the second derivative of a function, we need to differentiate it twice. Let's go through each function and find their second derivatives.
a) y = e^x + sin^2(x)
Step 1: Find the first derivative.
To find the first derivative of y, we will differentiate each term separately using the power rule and chain rule:
dy/dx = d/dx(e^x) + d/dx(sin^2(x))
Using the power rule, the derivative of e^x is e^x.
The derivative of sin^2(x) can be obtained using the chain rule. Let u = sin(x), then sin^2(x) = u^2. Therefore, our derivative becomes:
dy/dx = e^x + 2sin(x)cos(x)
Step 2: Find the second derivative.
To find the second derivative, we need to differentiate the derivative we found in step 1:
d^2y/dx^2 = d/dx(e^x + 2sin(x)cos(x))
The derivative of e^x is still e^x.
To find the derivative of 2sin(x)cos(x), we can use the product rule:
d/dx(2sin(x)cos(x)) = 2[cos(x)(cos(x))+sin(x)(-sin(x))]
Simplifying and rearranging, we get:
d^2y/dx^2 = e^x - 2sin^2(x)+2cos^2(x)
Therefore, the second derivative of y = e^x + sin^2(x) is d^2y/dx^2 = e^x - 2sin^2(x)+2cos^2(x).
b) y = 2√(x) + ln(x^3)
Step 1: Find the first derivative.
To find the first derivative of y, we will differentiate each term separately using the power rule and logarithmic differentiation:
dy/dx = d/dx(2√(x)) + d/dx(ln(x^3))
For 2√(x), we can rewrite it as 2x^(1/2) and then use the power rule to differentiate it:
d/dx(2x^(1/2)) = (1/2)*2*x^(-1/2) = x^(-1/2)
For ln(x^3), we can use logarithmic differentiation:
d/dx(ln(x^3)) = (1/x)*3x^2 = 3/x
Therefore, the first derivative is:
dy/dx = x^(-1/2) + 3/x
Step 2: Find the second derivative.
To find the second derivative, we need to differentiate the derivative found in step 1:
d^2y/dx^2 = d/dx(x^(-1/2) + 3/x)
The derivative of x^(-1/2) can be found using the power rule:
d/dx(x^(-1/2)) = (-1/2)*x^(-3/2)
The derivative of 3/x can be found using the power rule and quotient rule:
d/dx(3/x) = (3/x^2)*(-1/x) = -3/x^3
Therefore, the second derivative is:
d^2y/dx^2 = (-1/2)*x^(-3/2) - 3/x^3
So, the second derivative of y = 2√(x) + ln(x^3) is d^2y/dx^2 = (-1/2)*x^(-3/2) - 3/x^3.