find y' and y''
y = ln(x+(square root of (9+x^2))
is the square root in parenthesis within the ln? on one side it is and on the other its not
I figure it's
y = ln (x+√(9+x^2))
y' = 1/(x+√(9+x^2)) (1 + x/√(9+x^2))
= 1/√(9+x^2)
y" = -x/(9+x^2)^(3/2)
To find the first derivative (y') and the second derivative (y'') of the given function y = ln(x+√(9+x^2)), we can use the chain rule and the power rule.
First, let's rewrite the function using exponential notation: y = ln[(x+(9+x^2)^0.5)].
Now, let's find the first derivative (y'):
Step 1: Recall that the derivative of ln(u) is du/u.
Step 2: Apply the chain rule by multiplying the derivative of the inside function by the derivative of the inside function.
So, the derivative of (x+(9+x^2)^0.5) is 1 + (0.5)*(2x)(9+x^2)^(-0.5).
Step 3: Substitute the derivative of the inside function back into the original expression:
y' = (1 + (0.5)*(2x)(9+x^2)^(-0.5))/(x+(9+x^2)^0.5)
The first derivative y' is given by:
y' = (1 + x(9+x^2)^(-0.5))/(x+(9+x^2)^0.5)
To find the second derivative (y''), we will need to differentiate y' with respect to x.
Step 1: Differentiate the numerator and denominator separately using the quotient rule.
Step 2: Apply the chain rule again to find the derivative of (x+(9+x^2)^0.5) and (9+x^2)^(-0.5).
After simplifying the expression, the final expression for y'' is:
y'' = [(9+x^2)^0.5 - x(9+x^2)^(-0.5)] / (x+(9+x^2)^0.5)^2
Therefore, the first derivative y' is given by (1 + x(9+x^2)^(-0.5))/(x+(9+x^2)^0.5), and the second derivative y'' is [(9+x^2)^0.5 - x(9+x^2)^(-0.5)] / (x+(9+x^2)^0.5)^2.