Find the first and second derivatives of the given functions.
y=¡î(x©÷+1)
y'=x(x©÷+1)^(-1/2)
y"=[1(x©÷+1)^(-1/2)]-[(1/2)[(x©÷+1)^(-3/2)](2x)(x)]
y"=[(x©÷+1)^(-1/2)]-[(x©÷)(x©÷+1)^(-3/2)]
Is the second derivative right?
y=root(x^2+1)
y'=x(x^2+1)^(-1/2)
y"=[1(x^2+1)^(-1/2)]-[(1/2)[(x^2+1)^(-3/2)](2x)(x)]
y"=[(x^2+1)^(-1/2)]-[(x^2)(x^2+1)^(-3/2)]
I agree
To find the second derivative of the given function y = √(x^2 + 1), we first need to find the first derivative, and then differentiate it again.
Given the first derivative that you provided: y' = x(x^2 + 1)^(-1/2), we can differentiate this expression to find the second derivative.
To differentiate the function y' with respect to x, we can apply the chain rule. The chain rule states that if y = f(g(x)), then y' = f'(g(x)) * g'(x).
Let's apply the chain rule to find the second derivative:
y" = [x(x^2 + 1)^(-1/2)]'
= [x]' * (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]'
= 1 * (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]'
Now, let's differentiate the term x * [(x^2 + 1)^(-1/2)]' using the product rule. The product rule states that if y = f(x) * g(x), then y' = f'(x) * g(x) + f(x) * g'(x).
Let f(x) = x and g(x) = (x^2 + 1)^(-1/2). Using the product rule, we can differentiate the term x * [(x^2 + 1)^(-1/2)]:
[(x^2 + 1)^(-1/2)]' = (x)' * (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]'
Differentiating (x^2 + 1)^(-1/2), we can apply the chain rule again:
= 1 * (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]' * (x^2 + 1)
Now, we have the derivative. Let's substitute it back into the expression we previously obtained for the second derivative:
y" = 1 * (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]' * (x^2 + 1)
= (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]' * (x^2 + 1)
Therefore, the corrected expression for the second derivative (y") is:
y" = (x^2 + 1)^(-1/2) + x * [(x^2 + 1)^(-1/2)]' * (x^2 + 1)
Please note that this expression can be further simplified or rewritten if desired.