Find the 2nd Derivative f(x) = square root of x^2+1

Problem

ƒ(x)=√(,x^2+1)

Solution

1. Rewrite the function using a fractional exponent to prepare for differentiation.

ƒ(x)=(x^2+1)^(1/2)

2. Apply the power rule and chain rule to find the first derivative ƒ(x)^′

ƒ(x)^′=1/2*(x^2+1)^(−1/2)⋅(d(x^2)+1)/d(x)

3. Differentiate the inner function and simplify the expression.

ƒ(x)^′=1/2*(x^2+1)^(−1/2)⋅2*x

ƒ(x)^′=x*(x^2+1)^(−1/2)

4. Apply the product rule to find the second derivative ƒ(x)^″ where the rule is (d(u)*v)/d(x)=u^′*v+u*v^′

ƒ(x)^″=d(x)/d(x)*(x^2+1)^(−1/2)+x⋅d(x^2+1)/d(x)

5. Differentiate the terms using the power rule and chain rule again.

ƒ(x)^″=1⋅(x^2+1)^(−1/2)+x⋅(−1/2*(x^2+1)^(−3/2)⋅2*x)

6. Simplify the second term by multiplying the factors.

ƒ(x)^″=(x^2+1)^(−1/2)−x^2*(x^2+1)^(−3/2)

7. Factor out the common term (x^2+1)^(−3/2) to combine the expression.

ƒ(x)^″=(x^2+1)^(−3/2)⋅((x^2+1)^1−x^2)

8. Simplify the expression inside the parentheses.

ƒ(x)^″=(x^2+1)^(−3/2)⋅1

ƒ(x)^″=1/((x^2+1)^(3/2))

Final Answer

ƒ(x)^″=1/((x^2+1)√(,x^2+1))

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