How do you find the second derivative of # ln(x^2+5)# ?
Use natural logarithm derivative rules
For second derivative use the quotient rule
By signing up, you agree to our Terms of Service and Privacy Policy
To find the second derivative of ( \ln(x^2 + 5) ), follow these steps:

Find the first derivative using the chain rule: [ \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} ] [ u = x^2 + 5 ] [ \frac{du}{dx} = 2x ] [ \frac{d}{dx} \ln(x^2 + 5) = \frac{1}{x^2 + 5} \cdot 2x ] [ \frac{d}{dx} \ln(x^2 + 5) = \frac{2x}{x^2 + 5} ]

Find the second derivative using the quotient rule: [ \frac{d^2}{dx^2} \ln(u) = \frac{d}{dx} \left( \frac{1}{u} \cdot \frac{du}{dx} \right) ] [ = \frac{d}{dx} \left( \frac{2x}{x^2 + 5} \right) ] [ = \frac{(2)(x^2 + 5)  2x(2x)}{(x^2 + 5)^2} ] [ = \frac{2x^2 + 10  4x^2}{(x^2 + 5)^2} ] [ = \frac{2x^2 + 10}{(x^2 + 5)^2} ] [ = \frac{2(x^2  5)}{(x^2 + 5)^2} ]
Therefore, the second derivative of ( \ln(x^2 + 5) ) is ( \frac{2(x^2  5)}{(x^2 + 5)^2} ).
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7